A259754 -id:A259754

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A259748

a(n) = (Sum_{0<x<y<n} x*y) mod n.

+10
9

0, 0, 2, 3, 0, 1, 0, 2, 6, 0, 0, 5, 0, 7, 10, 4, 0, 12, 0, 15, 14, 11, 0, 22, 0, 0, 18, 21, 0, 5, 0, 8, 22, 0, 0, 15, 0, 19, 26, 10, 0, 28, 0, 33, 30, 23, 0, 44, 0, 0, 34, 39, 0, 9, 0, 14, 38, 0, 0, 25, 0, 31, 42, 16, 0, 44, 0, 51, 46, 35, 0, 66, 0, 0, 50

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

{a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.

From Danny Rorabaugh, Oct 22 2015: (Start)

a(n)/n = 0 iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);

a(n)/n = 1/6 iff n mod 24 = 6 (A259752);

a(n)/n = 1/4 iff n mod 24 = 8,16 (A259751);

a(n)/n = 5/12 iff n mod 24 = 12 (A073762);

a(n)/n = 1/2 iff n mod 24 = 14,22 (A259750);

a(n)/n = 2/3 iff n mod 24 = 3,9,15,18,21 (A259754);

a(n)/n = 3/4 iff n mod 24 = 4,20 (A259755);

a(n)/n = 11/12 iff n mod 24 = 0 (A008606).

(End)

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..24000

Danny Rorabaugh, Proof of a(n)/n values for A259748

FORMULA

a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.

a(24k) = 22k; a(24k+1) = 0; a(24k+2) = 0; a(24k+3) = 16k+2; a(24k+4) = 18k+3; a(24k+5) = 0; a(24k+6) = 4k+1, a(24k+7) = 0; a(24k+8) = 6k+2; a(24k+9) = 16k+6; a(24k+10) = 0; a(24k+11) = 0; a(24k+12) = 10k+5; a(24k+13) = 0; a(24k+14) = 12k+7; a(24k+15) = 16k+10; a(24k+16) = 6k+4; a(24k+17) = 0; a(24k+18) = 16k+12; a(24k+19) = 0; a(24k+20) = 18k+15; a(24k+21) = 16k+14; a(24k+22) = 12k+11; a(24k+23) = 0. - Danny Rorabaugh, Oct 22 2015

MATHEMATICA

A[n_]:=Sum[a b, {a, 1, n}, {b, a+1, n}]; Table[Mod[A[n], n], {n, 1, 122}]

PROG

(PARI) vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015

CROSSREFS

Cf. A000914,

A259749 (n such that a(n)=0),

A259750 (n such that n/a(n)=2),

A259751 (n such that n/a(n)=4),

A259752 (n such that n/a(n)=6),

A073762 (n such that n/a(n)=12/5),

A259754 (n such that n/a(n)=3/2),

A259755 (n such that n/a(n)=4/3),

A008606 (n such that n/a(n)=12/11).

KEYWORD

nonn,easy

AUTHOR

José María Grau Ribas, Jul 04 2015

STATUS

approved

A259755

Numbers that are congruent to {4, 20} mod 24.

+10
7

4, 20, 28, 44, 52, 68, 76, 92, 100, 116, 124, 140, 148, 164, 172, 188, 196, 212, 220, 236, 244, 260, 268, 284, 292, 308, 316, 332, 340, 356, 364, 380, 388, 404, 412, 428, 436, 452, 460, 476, 484, 500, 508, 524, 532, 548, 556, 572, 580, 596, 604, 620, 628

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

FORMULA

a(n) = 2*(6*n + (-1)^n - 3).

A259748(a(n))/a(n) = 3/4.

a(n) = 4*A007310(n). - Michel Marcus, Sep 22 2015

G.f.: 4*x*(1 + 4*x + x^2) / ((1 + x)*(1 - x)^2). - Bruno Berselli, Oct 23 2015

Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24. - Amiram Eldar, Dec 31 2021

E.g.f.: 2*(2 + (6*x - 3)*exp(x) + exp(-x)). - David Lovler, Sep 05 2022

MATHEMATICA

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 3/4 &]

Table[2 (6 n + (-1)^n - 3), {n, 1, 60}] (* Bruno Berselli, Oct 23 2015 *)

LinearRecurrence[{1, 1, -1}, {4, 20, 28}, 60] (* Harvey P. Dale, Jul 19 2016 *)

PROG

(Magma) [2*(6*n+(-1)^n-3): n in [1..60]]; // Vincenzo Librandi, Aug 27 2015

(PARI) vector(100, n, 2*(6*n+(-1)^n-3)) \\ Altug Alkan, Oct 23 2015

CROSSREFS

Cf. A000914, A007310.

Other sequences of numbers k such that A259748(k)/k equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259754.

KEYWORD

nonn,easy

AUTHOR

José María Grau Ribas, Jul 18 2015

EXTENSIONS

Better name from Danny Rorabaugh, Oct 22 2015

STATUS

approved

A259749

Numbers that are congruent to {1,2,5,7,10,11,13,17,19,23} mod 24.

+10
6

1, 2, 5, 7, 10, 11, 13, 17, 19, 23, 25, 26, 29, 31, 34, 35, 37, 41, 43, 47, 49, 50, 53, 55, 58, 59, 61, 65, 67, 71, 73, 74, 77, 79, 82, 83, 85, 89, 91, 95, 97, 98, 101, 103, 106, 107, 109, 113, 115, 119, 121, 122, 125, 127, 130, 131, 133, 137, 139, 143, 145

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

Original name: Numbers n such that A259748(n) = 0.

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-2,2,-1).

FORMULA

A259748(a(n)) = Sum_{x*y: x,y in Z/a(n)Z, x<>y} = 0.

G.f.: x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7) / ((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)). - Colin Barker, Aug 25 2016

MATHEMATICA

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #] == 0 & ]

Rest@ CoefficientList[Series[x (1 + x^2) (1 + 2 x^2 - x^3 + 2 x^4 - 2 x^5 + 3 x^6 + x^7)/((1 - x)^2*(1 - x + x^2 - x^3 + x^4) (1 + x + x^2 + x^3 + x^4)), {x, 0, 61}], x] (* Michael De Vlieger, Aug 25 2016 *)

Select[Range[150], MemberQ[{1, 2, 5, 7, 10, 11, 13, 17, 19, 23}, Mod[#, 24]]&] (* or *) LinearRecurrence[{2, -2, 2, -2, 2, -2, 2, -2, 2, -1}, {1, 2, 5, 7, 10, 11, 13, 17, 19, 23}, 70] (* Harvey P. Dale, Jan 15 2022 *)

PROG

(PARI) Vec(x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7)/((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

CROSSREFS

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259750, A259751, A259752, A259754, A259755.

KEYWORD

nonn,easy

AUTHOR

José María Grau Ribas, Jul 04 2015

EXTENSIONS

Better name from Danny Rorabaugh, Oct 22 2015

STATUS

approved

A259751

Numbers that are congruent to {8, 16} mod 24.

+10
6

8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136, 152, 160, 176, 184, 200, 208, 224, 232, 248, 256, 272, 280, 296, 304, 320, 328, 344, 352, 368, 376, 392, 400, 416, 424, 440, 448, 464, 472, 488, 496, 512, 520, 536, 544, 560, 568, 584, 592, 608, 616, 632

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Original name: Numbers n such that n/A259748(n) = 4.

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

FORMULA

A259748(a(n))/a(n) = 1/4.

a(n) = 8*A001651. - Danny Rorabaugh, Oct 22 2015

From Colin Barker, Aug 26 2016: (Start)

a(n) = 12*n-2*(-1)^n-6.

a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.

G.f.: 8*x*(1+x+x^2) / ((1-x)^2*(1+x)).

(End)

Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/72. - Amiram Eldar, Dec 31 2021

E.g.f.: 2*(4 + (6*x - 3)*exp(x) - exp(-x)). - David Lovler, Sep 05 2022

MATHEMATICA

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/4 & ]

PROG

(PARI) Vec(8*x*(1+x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

CROSSREFS

Cf. A000914, A001651.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259752, A259754, A259755.

KEYWORD

nonn,easy

AUTHOR

José María Grau Ribas, Jul 06 2015

EXTENSIONS

Better name from Danny Rorabaugh, Oct 22 2015

STATUS

approved

A259752

a(n) = 24*n - 18.

+10
6

6, 30, 54, 78, 102, 126, 150, 174, 198, 222, 246, 270, 294, 318, 342, 366, 390, 414, 438, 462, 486, 510, 534, 558, 582, 606, 630, 654, 678, 702, 726, 750, 774, 798, 822, 846, 870, 894, 918, 942, 966, 990, 1014, 1038, 1062, 1086, 1110, 1134, 1158, 1182, 1206

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Original name: Numbers n such that n/A259748(n) = 6.

Partial sums give A152746. - Leo Tavares, Jul 29 2023

LINKS

Danny Rorabaugh, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (2,-1).

FORMULA

A259748(a(n))/a(n) = 1/6.

a(n) = 6*A016813(n-1). - Michel Marcus, Jul 18 2015

G.f.: 6*x*(3*x+1)/(x-1)^2. - Alois P. Heinz, Jul 29 2023

MATHEMATICA

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/6 & ]

CROSSREFS

Cf. A000914, A063128, A063148, A152746.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259754, A259755.

KEYWORD

nonn,easy

AUTHOR

José María Grau Ribas, Jul 12 2015

EXTENSIONS

Better name from Danny Rorabaugh, Oct 22 2015

STATUS

approved

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