Displaying 11-17 of 17 results found.
4-step Fibonacci sequence starting with 1,0,1,0.
+0
6
1, 0, 1, 0, 2, 3, 6, 11, 22, 42, 81, 156, 301, 580, 1118, 2155, 4154, 8007, 15434, 29750, 57345, 110536, 213065, 410696, 791642, 1525939, 2941342, 5669619, 10928542, 21065442, 40604945, 78268548, 150867477, 290806412, 560547382, 1080489819
FORMULA
a(n+4) = a(n)+a(n+1)+a(n+2)+a(n+3).
PROG
(J) NB. see A251655 for the program and apply it to 1, 0, 1, 0.
CROSSREFS
Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251655, A251703, A251704, A251705.
4-step Fibonacci sequence starting with 0, 1, 1, 0.
+0
5
0, 1, 1, 0, 2, 4, 7, 13, 26, 50, 96, 185, 357, 688, 1326, 2556, 4927, 9497, 18306, 35286, 68016, 131105, 252713, 487120, 938954, 1809892, 3488679, 6724645, 12962170, 24985386, 48160880, 92833081, 178941517, 344920864, 664856342, 1281551804
FORMULA
a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
PROG
(J) NB. see A251655 for the program and apply it to 0, 1, 1, 0.
CROSSREFS
Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251655, A251656, A251672, A251703, A251704, A251705.
A Fielder sequence: a(n) = a(n-1) + a(n-2) + a(n-4).
(Formerly M2364 N0935)
+0
2
1, 3, 4, 11, 16, 30, 50, 91, 157, 278, 485, 854, 1496, 2628, 4609, 8091, 14196, 24915, 43720, 76726, 134642, 236283, 414645, 727654, 1276941, 2240878, 3932464, 6900996, 12110401, 21252275, 37295140, 65448411, 114853952, 201554638, 353703730, 620706779
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
G.f.: x*(1+2*x+4*x^3)/(1-x-x^2-x^4).
a(n) = n*Sum_{k=1..n} Sum_{j=floor((4*k-n)/3)..floor((4*k-n)/2)} binomial(j,n-4*k+3*j)*binomial(k,j))/k. - Vladimir Kruchinin, May 25 2011 a(n) = Trace(M^n), where M = [0, 0, 0, 1; 1, 0, 0, 0; 0, 1, 0, 1; 0, 0, 1, 1] is the companion matrix to the monic polynomial x^4 - x^3 - x^2 - 1. It follows that the sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for positive integers n and r and all primes p. See Zarelua. - Peter Bala, Dec 31 2022
MAPLE
A001641:=-(1+2*z+4*z**3)/(z+1)/(z**3-z**2+2*z-1); # conjectured by Simon Plouffe in his 1992 dissertation
MATHEMATICA
LinearRecurrence[{1, 1, 0, 1}, {1, 3, 4, 11}, 50] (* T. D. Noe, Aug 09 2012 *)
PROG
(PARI) a(n)=if(n<0, 0, polcoeff(x*(1+2*x+4*x^3)/(1-x-x^2-x^4)+x*O(x^n), n))
(Maxima) a(n):=(sum(sum(binomial(j, n-4*k+3*j)*binomial(k, j), j, floor((4*k-n)/3), floor((4*k-n)/2))/k, k, 1, n))*n; /* Vladimir Kruchinin, May 25 2011 */ (Magma) I:=[1, 3, 4, 11]; [n le 4 select I[n] else Self(n-1) + Self(n-2) + Self(n-4): n in [1..30]]; // G. C. Greubel, Jan 09 2018
A Fielder sequence.
(Formerly M2626 N1041)
+0
4
1, 3, 7, 11, 26, 45, 85, 163, 304, 578, 1090, 2057, 3888, 7339, 13862, 26179, 49437, 93366, 176321, 332986, 628852, 1187596, 2242800, 4235569, 7998951, 15106172, 28528288, 53876211, 101746240, 192149690, 362878313, 685302531, 1294206745, 2444133829
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
FORMULA
G.f.: x*(1+2*x+3*x^2+5*x^4)/(1-x-x^2-x^3-x^5).
a(n) = trace(M^n), where M = [0, 0, 0, 0, 1; 1, 0, 0, 0, 0; 0, 1, 0, 0, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1] is the 5 x 5 companion matrix to the monic polynomial x^5 - x^4 - x^3 - x^2 - 1. It follows that the sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for positive integers n and r and all primes p. See Zarelua. - Peter Bala, Jan 09 2023
MAPLE
A001645:=-(1+2*z+3*z**2+5*z**4)/(-1+z+z**2+z**3+z**5); [Conjectured by Simon Plouffe in his 1992 dissertation.]
MATHEMATICA
LinearRecurrence[{1, 1, 1, 0, 1}, {1, 3, 7, 11, 26}, 50] (* T. D. Noe, Aug 09 2012 *) CoefficientList[Series[x*(1+2*x+3*x^2+5*x^4)/(1-x-x^2-x^3-x^5), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
PROG
(PARI) a(n)=if(n<0, 0, polcoeff(x*(1+2*x+3*x^2+5*x^4)/(1-x-x^2-x^3-x^5)+x*O(x^n), n))
(Magma) I:=[1, 3, 7, 11, 26]; [n le 5 select I[n] else Self(n-1) + Self(n-2) + Self(n-3) + Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 19 2017
Array L(k,n) read by antidiagonals: k-step Lucas numbers.
+0
7
1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 1, 3, 7, 7, 1, 1, 3, 7, 11, 11, 1, 1, 3, 7, 15, 21, 18, 1, 1, 3, 7, 15, 26, 39, 29, 1, 1, 3, 7, 15, 31, 51, 71, 47, 1, 1, 3, 7, 15, 31, 57, 99, 131, 76, 1, 1, 3, 7, 15, 31, 63, 113, 191, 241, 123, 1
FORMULA
L(k,n) = L(k,n-1) + L(k,n-2) + ... + L(k,n-k); L(k,n) = -1 for n < 0, and L(k,0) = k.
G.f. for row k: x*(dB(k,x)/dx)/(1-B(k,x)), where B(k,x) = x + x^2 + ... + x^k. - Petros Hadjicostas, Jan 24 2019
EXAMPLE
Table begins:
1 | 1 1 1 1 1 1 1 1 1 1
2 | 1 3 4 7 11 18 29 47 76 123
3 | 1 3 7 11 21 39 71 131 241 443
4 | 1 3 7 15 26 51 99 191 367 708
5 | 1 3 7 15 31 57 113 223 439 863
6 | 1 3 7 15 31 63 120 239 475 943
7 | 1 3 7 15 31 63 127 247 493 983
8 | 1 3 7 15 31 63 127 255 502 1003
9 | 1 3 7 15 31 63 127 255 511 1013
PROG
(Sage)
def L(k, n):
if n < 0:
return -1
a = [-1]*(k-1) + [k] # [-1, -1, ..., -1, k]
for i in range(1, n+1):
a[:] = a[1:] + [sum(a)]
return a[-1]
[L(k, n) for d in (1..12) for k, n in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jan 10 2019
CROSSREFS
n-step Lucas number analog of A092921 Array F(k, n) read by antidiagonals: k-generalized Fibonacci numbers (and see related A048887, A048888). L(1, n) = "1-step Lucas numbers" = A000012. L(2, n) = 2-step Lucas numbers = A000204. L(3, n) = 3-step Lucas numbers = A001644. L(4, n) = 4-step Lucas numbers = A001648 Tetranacci numbers A073817 without the leading term 4. L(5, n) = 5-step Lucas numbers = A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15. L(6, n) = 6-step Lucas numbers = A074584 Esanacci ("6-anacci") numbers. L(7, n) = 7-step Lucas numbers = A104621 Heptanacci-Lucas numbers. L(8, n) = 8-step Lucas numbers = A105754. L(9, n) = 9-step Lucas numbers = A105755. See A000295, A125129 for comments on partial sums of diagonals. Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125129.
4-step Fibonacci sequence starting with 0, 1, 1, 1.
+0
5
0, 1, 1, 1, 3, 6, 11, 21, 41, 79, 152, 293, 565, 1089, 2099, 4046, 7799, 15033, 28977, 55855, 107664, 207529, 400025, 771073, 1486291, 2864918, 5522307, 10644589, 20518105, 39549919, 76234920, 146947533, 283250477, 545982849, 1052415779, 2028596638
FORMULA
a(n+4) = a(n) + a(n+1) + a(n+2) + a(n+3).
PROG
(J) (see www.jsoftware.com) First construct the generating matrix
[M=: (#.@}: + {:)\"1&.|: <:/~i.4
1 1 1 1
1 2 2 2
2 3 4 4
4 6 7 8
Given that matrix, one can produce the first 4*250 numbers with
, M(+/ . *)^:(i.250) 0 1 1 1x
CROSSREFS
Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251656, A251672, A251703, A251704, A251705.
Partial sums of diagonals of array of k-step Lucas numbers as in A125127, read by antidiagonals.
+0
2
1, 1, 4, 1, 8, 11, 1, 12, 19, 26, 1, 19, 33, 45, 57, 1, 30, 58, 84, 102, 120, 1, 48, 101, 157, 197, 222, 247, 1, 77, 179, 292, 380, 436, 469, 502, 1, 124, 318, 546, 731, 855, 929, 971, 1013, 1, 200, 567, 1026, 1409, 1674, 1838, 1932, 1984, 2036
COMMENTS
Array of partial sums of diagonals of L(k,n) begins: 0.|.1...4..11...26...57..120..247..502.1013.2036.
1.|.1...8..19...45..102..222..469..971.1984.
2.|.1..12..33...84..197..436..929.1932.
3.|.1..19..58..157..380..855.1838.
4.|.1..30.101..292..731.1674.
5.|.1..48.179..546.1409.
6.|.1..77.318.1026.
7.|.1.124.567.
8.|.1.200.
9.|.1.
FORMULA
Row 0 = SUM[i=1..n]L(i,i) = A127128 = partial sum of main diagonal of array of A125127. Row 1 = SUM[i=1..n]L(i,i+1) = partial sum of diagonal above main diagonal of array of A125127. Row 2 = SUM[i=1..n]L(i,i+2) = partial sum of diagonal 2 above main diagonal of array of A125127. .. Row m = SUM[i=1..n]L(i,i+m) = partial sum of diagonal 2 above main diagonal of array of A125127.
EXAMPLE
Row 1 of the derived array is the partial sum of the diagonal above the main diagonal of array of k-step Lucas numbers as in A125127, hence the partial sums of: 1, 7, 11, 26, 57, 120, 247, 502, 103, ... are 1 = 1; 8 = 1 + 7; 19 = 1 + 7 + 11; 45 = 1 + 7 + 11 + 26; and so forth.
CROSSREFS
Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125127, A000295.
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