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A386235 Number of partitions (p, q, r) of n into positive integers such that p + 11q + 13r is a perfect square.

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 6, 3, 7, 8, 6, 7, 6, 9, 8, 9, 9, 9, 15, 11, 13, 16, 13, 17, 15, 16, 17, 16, 19, 19, 23, 19, 21, 27, 23, 26, 24, 25, 29, 27, 28, 30, 32, 32, 34, 37, 35, 36, 37, 38, 40, 38, 38, 44, 46, 43, 46, 48, 50, 50, 48, 50, 50, 54, 52, 56, 60, 54, 64, 63, 62, 64

Offset: 3

Hoang Xuan Thanh, Jul 16 2025

nonn

For n >= 3 then there exists (p, q, r) | p + q + r = n such that p + 11*q + 13*r is a perfect square. This has been proven by Sylvester's theorem.

			n = 12: (4,4,4); 4 + 11*4 + 13*4 = 10^2; (7,4,1); 7 + 11*4 + 13*1 = 8^2; so a(12) = 2.

Hoang Xuan Thanh, Table of n, a(n) for n = 3..20000

a[n_]:=Module[{cnt=0,p,m2},Do[Do[p=n-q-r;m2=p +11*q+13*r;If[IntegerQ[Sqrt[m2]],cnt++],{r, 1, n - q - 1}],{q,1,n-2}];cnt];Array[a,78,3] (* James C. McMahon, Jul 22 2025 *)

a(n) = {my(cnt = 0); for (q = 1, n-2, for (r = 1, n - q - 1, p = n - q - r; m2 = p + 11*q + 13*r; if (issquare(m2), cnt++););); cnt;}

Conjecture: a(n) ~ K * n^(3/2) where K = 0.0914... from a(10000) = 91413 and a(20000) = 258667.

A380463 Partial sums of floor(n^2/13).

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 4, 7, 11, 17, 24, 33, 44, 57, 72, 89, 108, 130, 154, 181, 211, 244, 281, 321, 365, 413, 465, 521, 581, 645, 714, 787, 865, 948, 1036, 1130, 1229, 1334, 1445, 1562, 1685, 1814, 1949, 2091, 2239, 2394, 2556, 2725, 2902, 3086, 3278, 3478

Offset: 0

Hoang Xuan Thanh, Jun 22 2025

			a(9) = 0+0+0+0+1+1+2+3+4+6 = 17.

Hoang Xuan Thanh, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,0,0,0,0,0,0,1,-3,3,-1).

Cf. A173653, A173645, A175724.

LinearRecurrence[{3, -3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -3, 3,-1}, {0, 0, 0, 0, 1, 2, 4, 7, 11, 17, 24, 33, 44, 57, 72, 89}, 60]

a(n)=(2*n^3+3*n^2-35*n+48)\78 - ((n+6)%13<6)

(((x^4+x^9)*(1-x+x^2))/((1-x)^3*(1-x^13))).series(x, 52).coefficients(x, sparse=False) # Stefano Spezia, Jun 23 2025

G.f.: ((x^4+x^9)*(1-x+x^2))/((1-x)^3*(1-x^13)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-13) - 3*a(n-14) + 3*a(n-15) - a(n-16).
a(n) = floor((2*n^3 + 3*n^2 - 35*n + 48)/78) - [(n+6 mod 13)<6].

A384947 Positive integers m for which A183136(m) != f(m), where f(m) = floor( (m(m+1)/2)/phi - m/2 + 1/(2phi) ) and phi = (1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

15, 18, 36, 39, 41, 47, 49, 52, 91, 94, 96, 102, 103, 104, 107, 109, 123, 125, 128, 130, 136, 138, 141, 235, 238, 240, 246, 247, 248, 251, 252, 253, 267, 268, 269, 272, 273, 274, 277, 280, 281, 282, 285, 287, 303, 306, 322, 324, 327, 328

Offset: 1

Hoang Xuan Thanh, Jun 13 2025

nonn

f(m) is an approximation to A183136(m) = Sum_{k=1..m} floor(k/phi) based on assuming the floor in each term decreases it by 1/2 from what is otherwise a triangular sum; and further offset + 1/(2*phi) in f(m) chosen to improve the accuracy of this approximation.

The actual values of frac(k/phi) can differ from 1/2 each by a net amount which is enough to make m a term of this sequence.

			41 is term, because A183136(41) = 512 != 511 = floor(((41^2+1)*phi - 41) / (2*phi^2)).

Paolo Xausa, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Hoang Xuan Thanh)

Cf. A183136, A001622, A060143.

PositionIndex[MapIndexed[# != Floor[PolygonalNumber[#2[[1]]]/GoldenRatio - #2[[1]]/2 + 1/(2*GoldenRatio)] &, Accumulate[Floor[Range[500]/GoldenRatio]]]][True] (* Paolo Xausa, Jun 20 2025 *)

A384328 Expansion of 1 / ((1-x)^3 * (1-x^7)).

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 37, 48, 61, 76, 93, 112, 133, 157, 184, 214, 247, 283, 322, 364, 410, 460, 514, 572, 634, 700, 770, 845, 925, 1010, 1100, 1195, 1295, 1400, 1511, 1628, 1751, 1880, 2015, 2156, 2303, 2457, 2618, 2786, 2961

Offset: 0

Hoang Xuan Thanh, May 26 2025

Number of nonnegative integer solutions of equation x + y + z + 7*w=n.

a(n) is the number of partitions of n into parts 1 of three kinds and 7 (of one kind). - Joerg Arndt, May 28 2025

Hoang Xuan Thanh, Table of n, a(n) for n = 0..1999
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,1,-3,3,-1).

Cf. A000292, A002623, A014125, A122046, A122047, A175724.

my(x='x+O('x^50)); Vec(1/((1-x)^3*(1-x^7))) \\ Michel Marcus, May 27 2025

a(n) = ((n+2) * (n+9) * (n+4) - (r+2) * (r-5) * (r-3)) / 42 where r = n mod 7.

a(n) = floor((n+3) * (n^2+12*n+26) / 42).

A384063 Partial sums of A172471.

Original entry on oeis.org

0, 1, 3, 5, 7, 10, 13, 16, 20, 24, 28, 32, 36, 41, 46, 51, 56, 61, 67, 73, 79, 85, 91, 97, 103, 110, 117, 124, 131, 138, 145, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 233, 242, 251, 260, 269, 278, 287, 296, 305, 315, 325, 335, 345, 355, 365, 375, 385, 395, 405, 415, 426, 437

Offset: 0

Hoang Xuan Thanh, May 18 2025

nonn

Hoang Xuan Thanh, Table of n, a(n) for n = 0..10000

Cf. A172471, A173196, A000217.

Accumulate[Floor[Sqrt[2*Range[0, 100]]]] (* Paolo Xausa, Jun 04 2025 *)

a(n) = sum(k=1, n, sqrtint(2*k)); \\ Michel Marcus, May 23 2025

a(n) = m*n - floor((m-1)*(m+3)*(2m-1)/12), where m = A172471(n).

a(n) = m*n - A000217(m-1) - 2*A173196(m-1), where m = A172471(n).

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User: Hoang Xuan Thanh

A386235 Number of partitions (p, q, r) of n into positive integers such that p + 11*q + 13*r is a perfect square.

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A380463 Partial sums of floor(n^2/13).

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A384947 Positive integers m for which A183136(m) != f(m), where f(m) = floor( (m*(m+1)/2)/phi - m/2 + 1/(2*phi) ) and phi = (1+sqrt(5))/2 is the golden ratio.

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A384328 Expansion of 1 / ((1-x)^3 * (1-x^7)).

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A384063 Partial sums of A172471.

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A386235 Number of partitions (p, q, r) of n into positive integers such that p + 11q + 13r is a perfect square.

A384947 Positive integers m for which A183136(m) != f(m), where f(m) = floor( (m(m+1)/2)/phi - m/2 + 1/(2phi) ) and phi = (1+sqrt(5))/2 is the golden ratio.