Displaying 1-10 of 15 results found.
0, 1, 1, 4, 1, 5, 1, 6, 6, 7, 1, 1, 1, 9, 8, 8, 1, 1, 1, 3, 10, 13, 1, 1, 10, 15, 9, 1, 1, 1, 1, 10, 14, 19, 12, 10, 1, 21, 16, 1, 1, 1, 1, 3, 1, 25, 1, 1, 14, 3, 20, 1, 1, 1, 16, 1, 22, 31, 1, 4, 1, 33, 1, 12, 18, 1, 1, 3, 26, 1, 1, 12, 1, 39, 1, 1, 18, 1, 1, 1, 12, 43, 1, 2, 22, 45, 32, 1, 1, 1, 20, 3, 34, 49, 24
COMMENTS
For n >= 1, a(n) is a multiple of A373363(n).
PROG
(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
a(n) = gcd( A001414(n), A276085(n)), where A001414 is the sum of prime factors with repetition, and A276085 is the primorial base log-function.
+10
11
0, 1, 1, 2, 1, 1, 1, 3, 2, 7, 1, 1, 1, 1, 8, 4, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 2, 1, 12, 2, 1, 1, 8, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 4, 17, 1, 1, 8, 1, 2, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 6, 9, 1, 1, 4, 1, 1, 2, 2, 1, 8, 1, 1, 1, 20, 1, 2, 1, 12, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1
COMMENTS
As A001414 and A276085 are both fully additive sequences, all sequences that give the positions of multiples of some k > 1 in this sequence are closed under multiplication: For example, A373373, which gives the indices of multiples of 3.
PROG
(PARI)
A002110(n) = prod(i=1, n, prime(i));
A276085(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]* A002110(primepi(f[k, 1])-1)); };
CROSSREFS
Cf. A345452 (positions of even terms), A373373 (positions of multiples of 3).
a(n) = gcd( A001414(n), A083345(n)), where A001414 is the sum of prime factors with repetition, and A083345 is the numerator of the sum of the inverses of prime factors with repetition.
+10
10
0, 1, 1, 1, 1, 5, 1, 3, 2, 7, 1, 1, 1, 9, 8, 2, 1, 1, 1, 3, 10, 13, 1, 1, 2, 15, 1, 1, 1, 1, 1, 5, 14, 19, 12, 5, 1, 21, 16, 1, 1, 1, 1, 3, 1, 25, 1, 1, 2, 3, 20, 1, 1, 1, 16, 1, 22, 31, 1, 1, 1, 33, 1, 3, 18, 1, 1, 3, 26, 1, 1, 1, 1, 39, 1, 1, 18, 1, 1, 1, 4, 43, 1, 1, 22, 45, 32, 1, 1, 1, 20, 3, 34, 49, 24, 1, 1, 1, 1, 7
PROG
(PARI)
A083345(n) = { my(f=factor(n)); numerator(vecsum(vector(#f~, i, f[i, 2]/f[i, 1]))); };
CROSSREFS
Cf. A345452 (positions of even terms), A353374 (their characteristic function).
a(n) = gcd(sum of distinct prime factors of n, product of distinct prime factors of n).
+10
7
1, 2, 3, 2, 5, 1, 7, 2, 3, 1, 11, 1, 13, 1, 1, 2, 17, 1, 19, 1, 1, 1, 23, 1, 5, 1, 3, 1, 29, 10, 31, 2, 1, 1, 1, 1, 37, 1, 1, 1, 41, 6, 43, 1, 1, 1, 47, 1, 7, 1, 1, 1, 53, 1, 1, 1, 1, 1, 59, 10, 61, 1, 1, 2, 1, 2, 67, 1, 1, 14, 71, 1, 73, 1, 1, 1, 1, 6, 79, 1, 3, 1, 83, 6, 1, 1, 1, 1, 89, 10, 1, 1, 1
EXAMPLE
n=84: a(84) = gcd(2*3*7, 2+3+7) = gcd(42, 12) = 6.
MATHEMATICA
PrimeFactors[n_Integer] := Flatten[ Table[ # [[1]], {1}] & /@ FactorInteger[n]]; f[n_] := Block[{pf = PrimeFactors[n]}, GCD[Plus @@ pf, Times @@ pf]]; Table[ f[n], {n, 93}] (* Robert G. Wilson v, Nov 04 2004 *)
CROSSREFS
Differs from related A099635 for the first time at n=84, where a(84) = 6, while A099635(84) = 12.
Differs from A014963 for the first time at n=30, where a(30) = 10, while A014963(30) = 1.
0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 2, 4, 1, 1, 1, 3, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 2, 1, 2, 2, 1, 1, 2, 1, 1, 3, 1, 3, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 2, 1, 2, 1, 1, 4, 1, 1, 1, 6, 2, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 4, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 3, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3
PROG
(PARI)
A001414(n) = ((n=factor(n))[, 1]~*n[, 2]);
A059975(n) = {my(f = factor(n)); sum(i = 1, #f~, f[i, 2]*(f[i, 1] - 1)); };
3-Suzanne numbers; composite multiples of 3 whose sum of prime factors with multiplicity is a multiple of 3.
+10
6
9, 24, 27, 42, 60, 72, 78, 81, 105, 114, 126, 132, 150, 180, 186, 192, 195, 204, 216, 222, 231, 234, 243, 258, 276, 285, 315, 330, 336, 342, 348, 357, 366, 375, 378, 396, 402, 429, 438, 450, 465, 474, 480, 483, 492, 510, 540, 555, 558, 564, 576, 582, 585
COMMENTS
Composite numbers k such that the sum of digits of k ( A007953) and the sum of sums of digits of the prime factors of k (taken with multiplicity, A118503) are both divisible by 3. - Amiram Eldar, Apr 23 2021
The new secondary definition is equal to the original because taking the decimal digit sum preserves congruence modulo 3. This is a multiplicative semigroup: if m and n are in the sequence, then so is m*n. - Antti Karttunen, Jun 08 2024
REFERENCES
József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 384.
James J. Tattersall, Elementary Number Theory in Nine Chapters, 2nd ed., Cambridge University Press, 2005, p. 93.
EXAMPLE
42 = 2*3*7 is a term as it is a multiple of 3, and also 2+3+7 = 12 is a multiple of 3.
60 = 2*2*3*5 is a term is it is a multiple of 3, and also 2+2+3+5 = 12 is a multiple of 3.
(End)
MATHEMATICA
s[n_] := Plus @@ IntegerDigits[n]; f[p_, e_] := e*s[p]; sp[n_] := Plus @@ f @@@ FactorInteger[n]; suz3Q[n_] := CompositeQ[n] && And @@ Divisible[{s[n], sp[n]}, 3]; Select[Range[600], suz3Q] (* Amiram Eldar, Apr 23 2021 *)
PROG
(PARI) isA102217(n) = if(n<=3 || (n%3), 0, my(f=factor(n)); 0==(sum(i=1, #f~, f[i, 2]*sumdigits(f[i, 1]))%3)); \\ Antti Karttunen, Jun 08 2024
EXTENSIONS
Alternative definition added and keyword:base removed by Antti Karttunen, Jun 08 2024
a(n) = gcd(n, A059975(n)), where A059975 is fully additive with a(p) = p-1.
+10
6
1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 1, 4, 1, 7, 3, 4, 1, 1, 1, 2, 1, 11, 1, 1, 1, 13, 3, 4, 1, 1, 1, 1, 3, 17, 5, 6, 1, 19, 1, 1, 1, 3, 1, 4, 1, 23, 1, 6, 1, 1, 3, 2, 1, 1, 1, 1, 1, 29, 1, 4, 1, 31, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 1, 37, 5, 4, 1, 3, 1, 8, 1, 41, 1, 2, 5, 43, 3, 1, 1, 9, 1, 4, 1, 47, 1, 1, 1, 1, 1, 10
PROG
(PARI)
A059975(n) = {my(f = factor(n)); sum(i = 1, #f~, f[i, 2]*(f[i, 1] - 1)); };
Numerator of sopfr(n)/n, where sopfr= A001414 is the sum of prime factors (with repetition).
+10
4
0, 1, 1, 1, 1, 5, 1, 3, 2, 7, 1, 7, 1, 9, 8, 1, 1, 4, 1, 9, 10, 13, 1, 3, 2, 15, 1, 11, 1, 1, 1, 5, 14, 19, 12, 5, 1, 21, 16, 11, 1, 2, 1, 15, 11, 25, 1, 11, 2, 6, 20, 17, 1, 11, 16, 13, 22, 31, 1, 1, 1, 33, 13, 3, 18, 8, 1, 21, 26, 1, 1, 1, 1, 39, 13, 23, 18, 3, 1, 13, 4, 43, 1, 1, 22, 45, 32, 17
EXAMPLE
n=200: (2+2+2+5+5)/(2*2*2*5*5) = 16/(2*2*2*5*5) = (2*2*2*2)/(2*2*2*5*5) = 2/25, therefore a(200)=2, A082344(200)=25.
MATHEMATICA
sopfr[n_] := If[n == 1, 0, Total[Times @@@ FactorInteger[n]]];
a[n_] := Numerator[sopfr[n]/n];
Denominator of sopfr(n)/n, where sopfr= A001414 is the sum of prime factors (with repetition).
+10
4
1, 1, 1, 1, 1, 6, 1, 4, 3, 10, 1, 12, 1, 14, 15, 2, 1, 9, 1, 20, 21, 22, 1, 8, 5, 26, 3, 28, 1, 3, 1, 16, 33, 34, 35, 18, 1, 38, 39, 40, 1, 7, 1, 44, 45, 46, 1, 48, 7, 25, 51, 52, 1, 54, 55, 56, 57, 58, 1, 5, 1, 62, 63, 16, 65, 33, 1, 68, 69, 5, 1, 6, 1, 74, 75, 76, 77, 13, 1, 80, 27, 82, 1, 6
EXAMPLE
n=200: (2+2+2+5+5)/(2*2*2*5*5) = 16/(2*2*2*5*5) = (2*2*2*2)/(2*2*2*5*5) = 2/25, therefore a(200)=25, A082343(200)=2.
MATHEMATICA
sopd[n_]:=Module[{f=Flatten[Table[#[[1]], #[[2]]]&/@FactorInteger[n]]}, Denominator[ Total[f]/n]]; Array[sopd, 90] (* Harvey P. Dale, Jul 24 2018 *)
sopfr[n_] := If[n == 1, 0, Total[Times @@@ FactorInteger[n]]];
a[n_] := Denominator[sopfr[n]/n];
a(0)=1; for n > 0, a(n) = the greatest common divisor (GCD) of n and the sum of all previous terms if the GCD is not already in the sequence; otherwise a(n) = a(n-1) + n.
+10
4
1, 2, 4, 7, 11, 5, 6, 13, 21, 30, 10, 21, 33, 46, 14, 29, 45, 62, 18, 37, 57, 78, 22, 45, 69, 94, 26, 53, 81, 110, 140, 171, 203, 236, 270, 305, 341, 378, 416, 39, 79, 120, 162, 205, 249, 294, 340, 387, 3, 52, 102, 17, 69, 122, 176, 231, 287, 344, 402, 461, 521, 582, 644, 707, 771, 836, 902, 969
COMMENTS
The sequence displays the unusual behavior of decreasing 53 times in the first 1975 terms, due to the existence of a GCD which has not previously appeared in the sequence, but then not decreasing again for n up to at least 100 million. In this period there are 37 repeated terms, the first being 21 at n=11 and the last 161202 at n=2054. In the same range many values do not appear, for example 16,23,28,32,36. It is unknown when the sequence decreases again, or if all values eventually appear. The 100 millionth term is 4999999948050717.
See the companion sequence A333980 for the sum of the terms from a(0) to a(n).
EXAMPLE
a(2) = 4 as the sum of all previous terms is a(0)+a(1) = 3, and the GCD of 3 and 2 is 1, which has already appeared in the sequence. Therefore a(2) = a(1) + n = 2 + 2 = 4.
a(4) = 11 as the sum of all previous terms is a(0)+...+a(3) = 14, and the GCD of 14 and 4 is 2. However 2 has already appeared so a(4) = a(3) + n = 7 + 4 = 11.
a(5) = 5 as the sum of all previous terms is a(0)+...+a(4) = 25, and the GCD of 25 and 5 is 5, and as 5 has not previous appeared a(5) = 5.
PROG
(PARI) lista(nn) = {my(va = vector(nn), s=0); va[1] = 1; s += va[1]; for (n=2, nn, my(g = gcd(n-1, s)); if (#select(x->(x==g), va), va[n] = va[n-1]+n-1, va[n] = g); s += va[n]; ); va; } \\ Michel Marcus, Sep 05 2020
Search completed in 0.012 seconds
|