cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A294654 Expansion of Product_{k>=1} 1/((1 - x^(2*k-1))^(k*(5*k-3)/2)*(1 - x^(2*k))^(k*(5*k+3)/2)).

Original entry on oeis.org

1, 1, 5, 12, 35, 81, 208, 475, 1123, 2505, 5617, 12192, 26368, 55797, 117255, 242660, 498126, 1010515, 2033662, 4053214, 8017622, 15729219, 30643069, 59268267, 113898873, 217480476, 412813600, 779042099, 1462188257, 2729852845, 5070966794, 9373909586, 17247473718
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 06 2017

Keywords

Comments

Euler transform of the generalized heptagonal numbers (A085787).

Crossrefs

Programs

  • Mathematica
    nmax = 32; CoefficientList[Series[Product[1/((1 - x^(2 k - 1))^(k (5 k - 3)/2) (1 - x^(2 k))^(k (5 k + 3)/2)), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d (5 d (d + 1)/8 + (-1)^d (2 d + 1)/16 - 1/16), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 32}]

Formula

G.f.: Product_{k>=1} 1/(1 - x^k)^A085787(k).
a(n) ~ exp(Pi * (2/3)^(5/4) * n^(3/4) + 5*Zeta(3) * sqrt(3*n) / (2^(3/2) * Pi^2) - (75*3^(1/4) * Zeta(3)^2 / (2^(13/4) * Pi^5) + Pi / (2^(17/4) * 3^(3/4))) * n^(1/4) + 375 * Zeta(3)^3 / (8*Pi^8) - 5*Zeta(3) / (64*Pi^2) + 1/12) * Pi^(1/12) / (A * 2^(11/6) * 3^(7/48) * n^(31/48)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 07 2017

A294840 Expansion of Product_{k>=1} (1 + x^(2*k-1))^(k*(5*k-3)/2)*(1 + x^(2*k))^(k*(5*k+3)/2).

Original entry on oeis.org

1, 1, 4, 11, 26, 65, 150, 343, 760, 1670, 3574, 7561, 15752, 32396, 65850, 132386, 263447, 519316, 1014744, 1966234, 3780464, 7215020, 13674227, 25744768, 48166429, 89576421, 165638008, 304615115, 557275053, 1014398476, 1837617957, 3313527482, 5948262037, 10632231253, 18926026208
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 09 2017

Keywords

Comments

Weigh transform of the generalized heptagonal numbers (A085787).

Crossrefs

Programs

  • Mathematica
    nmax = 34; CoefficientList[Series[Product[(1 + x^(2 k - 1))^(k (5 k - 3)/2) (1 + x^(2 k))^(k (5 k + 3)/2), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d (5 d (d + 1)/8 + (-1)^d (2 d + 1)/16 - 1/16), {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 34}]

Formula

G.f.: Product_{k>=1} (1 + x^k)^A085787(k).
a(n) ~ 7^(1/8) * exp(Pi*sqrt(2) * 7^(1/4) * n^(3/4) / 3^(5/4) + 15*Zeta(3) * sqrt(3*n/7) / (4*Pi^2) - (7*Pi^6 + 4050*Zeta(3)^2)*n^(1/4) / (112*sqrt(2) * 3^(3/4) * 7^(1/4) * Pi^5) + 15*Zeta(3) * (7*Pi^6 + 5400*Zeta(3)^2) / (3136*Pi^8)) / (2^(7/3) * 3^(1/8) * n^(5/8)). - Vaclav Kotesovec, Nov 10 2017

A317300 Sequence obtained by taking the general formula for generalized k-gonal numbers: m*((k - 2)*m - k + 4)/2, where m = 0, +1, -1, +2, -2, +3, -3, ... and k >= 5. Here k = 0.

Original entry on oeis.org

0, 1, -3, 0, -8, -3, -15, -8, -24, -15, -35, -24, -48, -35, -63, -48, -80, -63, -99, -80, -120, -99, -143, -120, -168, -143, -195, -168, -224, -195, -255, -224, -288, -255, -323, -288, -360, -323, -399, -360, -440, -399, -483, -440, -528, -483, -575, -528, -624, -575, -675, -624, -728, -675, -783
Offset: 0

Views

Author

Omar E. Pol, Jul 29 2018

Keywords

Comments

Taking the same formula with k = 1 we have A317301.
Taking the same formula with k = 2 we have A001057 (canonical enumeration of integers).
Taking the same formula with k = 3 we have 0 together with A008795 (Molien series for 3-dimensional representation of dihedral group D_6 of order 6).
Taking the same formula with k = 4 we have A008794 (squares repeated) except the initial zero.
Taking the same formula with k >= 5 we have the generalized k-gonal numbers (see Crossrefs section).

Crossrefs

Row 0 of A303301.
Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

Programs

  • PARI
    concat(0, Vec(x*(1 - 4*x + x^2) / ((1 - x)^3*(1 + x)^2) + O(x^50))) \\ Colin Barker, Aug 01 2018

Formula

a(n) = -A174474(n+1).
From Colin Barker, Aug 01 2018: (Start)
G.f.: x*(1 - 4*x + x^2) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = -n*(n + 4) / 4 for n even.
a(n) = -(n - 3)*(n + 1) / 4 for n odd.
(End)

A317301 Sequence obtained by taking the general formula for generalized k-gonal numbers: m*((k - 2)*m - k + 4)/2, where m = 0, +1, -1, +2, -2, +3, -3, ... and k >= 5. Here k = 1.

Original entry on oeis.org

0, 1, -2, 1, -5, 0, -9, -2, -14, -5, -20, -9, -27, -14, -35, -20, -44, -27, -54, -35, -65, -44, -77, -54, -90, -65, -104, -77, -119, -90, -135, -104, -152, -119, -170, -135, -189, -152, -209, -170, -230, -189, -252, -209, -275, -230, -299, -252, -324, -275, -350, -299, -377, -324, -405, -350, -434
Offset: 0

Views

Author

Omar E. Pol, Jul 29 2018

Keywords

Comments

Taking the same formula with k = 0 we have A317300.
Taking the same formula with k = 2 we have A001057 (canonical enumeration of integers).
Taking the same formula with k = 3 we have 0 together with A008795 (Molien series for 3-dimensional representation of dihedral group D_6 of order 6).
Taking the same formula with k = 4 we have A008794 (squares repeated) except the initial zero.
Taking the same formula with k >= 5 we have the generalized k-gonal numbers (see Crossrefs section).

Crossrefs

Row 1 of A303301.
Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

Programs

  • Magma
    /* By definition: */ k:=1; [0] cat [m*i*((k-2)*m*i-k+4)/2: i in [1,-1], m in [1..30]]; // Bruno Berselli, Jul 30 2018
    
  • Mathematica
    Table[(-2 n (n + 1) - 5 (2 n + 1) (-1)^n + 5)/16, {n, 0, 60}] (* Bruno Berselli, Jul 30 2018 *)
  • PARI
    concat(0, Vec(x*(1 - 3*x + x^2)/((1 + x)^2*(1 - x)^3) + O(x^50))) \\ Colin Barker, Aug 01 2018

Formula

From Bruno Berselli, Jul 30 2018: (Start)
O.g.f.: x*(1 - 3*x + x^2)/((1 + x)^2*(1 - x)^3).
E.g.f.: (-5*(1 + 2*x) + (5 - 2*x^2)*exp(2*x))*exp(-x)/16.
a(n) = a(-n+1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = (-2*n*(n + 1) - 5*(2*n + 1)*(-1)^n + 5)/16. Therefore:
a(n) = -n*(n + 6)/8 for even n;
a(n) = -(n - 5)*(n + 1)/8 for odd n. Also:
a(n) = a(n-5) for odd n > 3.
2*(2*n - 1)*a(n) + 2*(2*n + 1)*a(n-1) + n*(n^2 - 3) = 0. (End)

A195018 a(n) = n*(10*n-3).

Original entry on oeis.org

0, 7, 34, 81, 148, 235, 342, 469, 616, 783, 970, 1177, 1404, 1651, 1918, 2205, 2512, 2839, 3186, 3553, 3940, 4347, 4774, 5221, 5688, 6175, 6682, 7209, 7756, 8323, 8910, 9517, 10144, 10791, 11458, 12145, 12852, 13579, 14326, 15093, 15880, 16687, 17514, 18361, 19228
Offset: 0

Views

Author

Omar E. Pol, Oct 14 2011

Keywords

Comments

Bisection of heptagonal numbers A000566.
Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. This is one of the four semi-diagonals of the square spiral.
Also sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. Semi-axis perpendicular to the main axis A195015 in the same spiral.

Crossrefs

Programs

Formula

a(n) = A153127(n-1) + 1, if n >= 1.
G.f.: -x*(7+13*x)/(x-1)^3. - R. J. Mathar, Oct 15 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=7, a(2)=34. - Harvey P. Dale, May 27 2012
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: exp(x)*x*(7 + 10*x).
a(n) = A000566(2*n). (End)

A253187 Number of ordered ways to write n as the sum of a pentagonal number, a second pentagonal number and a generalized decagonal number.

Original entry on oeis.org

1, 2, 2, 2, 1, 1, 1, 3, 4, 2, 2, 1, 4, 3, 3, 4, 2, 3, 1, 3, 2, 2, 5, 3, 3, 3, 3, 6, 3, 6, 4, 2, 3, 1, 7, 2, 4, 5, 5, 4, 1, 5, 5, 2, 3, 4, 4, 5, 5, 5, 3, 5, 7, 6, 4, 3, 1, 6, 6, 8, 5, 3, 6, 4, 7, 4, 2, 6, 5, 5, 3, 4, 8, 3, 3, 3, 6, 6, 7, 9, 6, 2, 5, 6, 7, 7, 4, 6, 6, 7, 5, 3, 10, 6, 3, 4, 5, 7, 3, 10, 7
Offset: 0

Views

Author

Zhi-Wei Sun, Apr 07 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n. Also, for any ordered pair (k,m) among (5,7), (5,9), (5,13), (6,5), (6,7), (7,5), each nonnegative integer n can be written as the sum of a k-gonal number, a second k-gonal number and a generalized m-gonal number.
See also the author's similar conjectures in A254574, A254631, A255916 and the two linked papers.

Examples

			a(33) = 1 since 33 = 0*(3*0-1)/2 + 4*(3*4+1)/2 + 1*(4*1+3).
a(56) = 1 since 56 = 4*(3*4-1)/2 + 2*(3*2+1)/2 + 3*(4*3+3).
		

Crossrefs

Programs

  • Mathematica
    DQ[n_]:=IntegerQ[Sqrt[16n+9]]
    Do[r=0;Do[If[DQ[n-x(3x-1)/2-y(3y+1)/2],r=r+1],{x,0,(Sqrt[24n+1]+1)/6},{y,0,(Sqrt[24(n-x(3x-1)/2)+1]-1)/6}];
    Print[n," ",r];Continue,{n,0,100}]

A255916 Number of ways to write n as the sum of a generalized heptagonal number, an octagonal number and a nonagonal number.

Original entry on oeis.org

1, 3, 3, 1, 1, 2, 1, 1, 3, 4, 3, 1, 1, 3, 3, 2, 2, 2, 2, 2, 1, 3, 4, 2, 2, 3, 3, 3, 5, 3, 2, 2, 2, 1, 3, 5, 4, 3, 1, 2, 2, 2, 3, 4, 3, 3, 3, 5, 5, 3, 3, 3, 2, 3, 4, 5, 5, 2, 4, 4, 1, 1, 1, 3, 5, 4, 3, 6, 4, 1, 3, 5, 5, 2, 4, 3, 5, 3, 4, 6, 5, 4, 4, 5, 2, 2, 2, 6, 2, 3, 5, 4, 4, 5, 3, 3, 5, 3, 3, 3, 8
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 11 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n. Moreover, for k >= j >=3, every nonnegative integer can be written as the sum of a generalized heptagonal number, a j-gonal number and a k-gonal number, if and only if (j,k) is among the following ordered pairs:
(3,k) (k = 3..19, 21..24, 26, 27, 29, 30), (4,k) (k = 4..11, 13, 14, 17, 19, 20, 23, 26), (5,6), (5,9), (6,7), (8,9).
(ii) For k >= j >= 3, every nonnegative integer can be written as the sum of a generalized pentagonal number, a j-gonal number and a k-gonal number, if and only if (j,k) is among the following ordered pairs:
(3,k) (k = 3..20, 22, 24, 25, 28..30, 32, 37), (4,k) (k = 4..13, 15, 16, 18, 20..25, 27, 28, 31, 33, 34), (5,k) (k = 6..12, 20), (6,k) (k = 7..10), (7,9), (7,11), (8,10), (9,11).

Examples

			 a(60) = 1 since 60 = (-2)(5*(-2)-3)/2 + 1*(3*1-2) + 4*(7*4-5)/2.
a(279) = 1 since 279 = 3*(5*3-3)/2 + 0*(3*0-2) + 9*(7*9-5)/2.
		

Crossrefs

Programs

  • Mathematica
    HQ[n_]:=HQ[n]=IntegerQ[Sqrt[40n+9]]&&(Mod[Sqrt[40n+9]+3,10]==0||Mod[Sqrt[40n+9]-3,10]==0)
    Do[r=0;Do[If[HQ[n-x(3x-2)-y(7y-5)/2],r=r+1],{x,0,(Sqrt[3n+1]+1)/3},{y,0,(Sqrt[56(n-x(3x-2))+25]+5)/14}];
    Print[n," ",r];Continue,{n,0,100}]

A256171 Number of ways to write n as the sum of three unordered generalized heptagonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 0, 1, 2, 1, 2, 3, 0, 1, 3, 1, 2, 3, 1, 1, 1, 1, 3, 3, 1, 2, 1, 2, 3, 1, 2, 4, 2, 1, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 3, 2, 1, 2, 2, 3, 5, 2, 2, 2, 2, 3, 4, 2, 2, 4, 1, 3, 2, 1, 4, 3, 2, 2, 5, 2, 4, 3, 0, 4, 2, 1, 3, 6, 3, 3, 3, 1, 5, 2, 3, 5, 2, 2, 3, 3, 1, 5, 3, 1, 3, 3, 4
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 17 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 10, 16, 76, 307.
(ii) For any integer m > 2 not divisible by 4, each sufficiently large integer n can be written as the sum of three generalized m-gonal numbers.
In 1994 R. K. Guy noted that none of 10, 16 and 76 can be written as the sum of three generalized heptagonal numbers.

Examples

			a(157) = 1 since 157 = 3*(5*3-3)/2 + (-3)*(5*(-3)-3)/2 + 7*(5*7-3)/2.
a(748) = 1 since 748 = 0*(5*0-3)/2 + 0*(5*0-3)/2 + (-17)*(5*(-17)-3)/2.
		

Crossrefs

Programs

  • Mathematica
    T[n_]:=Union[Table[x(5x-3)/2, {x, -Floor[(Sqrt[40n+9]-3)/10], Floor[(Sqrt[40n+9]+3)/10]}]]
    L[n_]:=Length[T[n]]
    Do[r=0;Do[If[Part[T[n],x]>n/3,Goto[aa]];Do[If[Part[T[n],x]+2*Part[T[n],y]>n,Goto[bb]];
    If[MemberQ[T[n], n-Part[T[n],x]-Part[T[n],y]]==True,r=r+1];
    Continue,{y,x,L[n]}];Label[bb];Continue,{x,1,L[n]}];Label[aa];Print[n," ",r];Continue, {n,0,100}]

A270705 Number of ordered ways to write n as x^2*pen(x) + pen(y) + pen(z) with pen(x) = x*(3x+1)/2 and pen(y) <= pen(z), where x, y and z are integers ("pen" stands for "pentagonal").

Original entry on oeis.org

1, 2, 5, 5, 6, 4, 3, 4, 4, 4, 3, 2, 3, 3, 6, 4, 4, 4, 3, 3, 3, 4, 6, 5, 6, 5, 5, 8, 8, 9, 7, 5, 7, 6, 7, 9, 7, 10, 5, 5, 9, 6, 12, 7, 8, 6, 3, 10, 6, 5, 7, 5, 8, 7, 8, 9, 5, 9, 8, 7, 5, 7, 7, 5, 6, 6, 5, 4, 6, 4, 8, 5, 9, 6, 3, 7, 5, 8, 8, 8, 8, 6, 6, 6, 6, 6, 8, 3, 1, 4, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 21 2016

Keywords

Comments

Conjecture: (i) Any natural number can be written as a*x^2*pen(x) + b*pen(y) + c*pen(z) with x, y and z integers, provided that (a,b,c) is among the following ordered triples: (j,1,k) (j = 1,2; k = 1,2,3,4), (1,2,3), (3,1,4) and (4,1,3).
(ii) Every n = 0,1,2,... can be expressed as x^2*pen(x) + T(y) + T(z) with x, y and z integers, where T(m) denotes the triangular number m*(m+1)/2. Also, for each (a,b) = (1,2),(1,4),(2,2), any natural number can be written as a*x^2*T(x) + b*T(y) + T(z) with x, y and z integers.
(iii) Each natural number can be written as x^2*P(x) + pen(y) + pen(z) with x, y and z integers, where P(x) is either of the following polynomials: a*T(x) (a = 1,2,3,4,5), x*(5x+3)/2, x*(3x+1), x*(3x+2), x*(7x+1)/2, x*(4x+1), x*(4x+3), x*(9x+5)/2, x*(5x+3), x*(11x+9)/2, x*(13x+5)/2, x*(17x+9)/2, 3x*(3x+2), x*(11x+2).
See also A270594 and A270706 for other similar conjectures.

Examples

			a(88) = 1 since 88 = 1^2*pen(1) + pen(-5) + pen(-6).
		

Crossrefs

Programs

  • Mathematica
    pen[x_]:=pen[x]=x(3x+1)/2
    pQ[n_]:=pQ[n]=IntegerQ[Sqrt[24n+1]]
    Do[r=0;Do[If[pQ[n-pen[y]-x^2*pen[x]],r=r+1],{y,-Floor[(Sqrt[12n+1]+1)/6],(Sqrt[12n+1]-1)/6},{x,-1-Floor[(2(n-pen[y])/3)^(1/4)],(2(n-pen[y])/3)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]

A270706 Number of ordered ways to write n as x^2*T(x) + y^2 + T(z), where x, y and z are integers with x nonzero, y positive and z nonnegative, and T(m) denotes the triangular number m*(m+1)/2.

Original entry on oeis.org

1, 2, 1, 2, 4, 2, 2, 4, 2, 3, 5, 2, 3, 4, 3, 6, 4, 2, 6, 5, 2, 4, 6, 2, 3, 7, 3, 5, 6, 4, 8, 5, 2, 5, 3, 5, 9, 7, 3, 5, 8, 3, 6, 5, 2, 8, 4, 2, 9, 6, 4, 7, 7, 4, 5, 7, 5, 9, 5, 3, 7, 4, 5, 12, 9, 4, 5, 8, 4, 6, 11, 3, 9, 5, 3, 10, 3, 4, 9, 6, 5, 11, 8, 5, 7, 9, 3, 5, 4, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 21 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 90, 438, 480, 7108.
(ii) Let pen(x) = x*(3x+1)/2. Any natural number can be written as a*f(x)*g(x) + f(y) + g(z) with x, y and z integers, whenever (a,f(x),g(x)) is among the following ordered triples: (1,T(x),x^2), (1,T(x),pen(x)), (1,T(x),x*(5x+1)/2), (1,T(x),x*(5x+3)/2), (1,T(x),x*(3x+j)) (j = 1,2), (1,pen(x),3*T(x)), (1,pen(x),x*(7x+j)/2) (j = 1,3,5), (1,pen(x),x*(4x+1)), (2,T(x),x^2), (2,T(x),pen(x)), (2,T(x),x(5x+j)/2) (j = 1,3), (2,T(x),x*(3x+j)) (j = 1,2), (2,2*T(x),pen(x)), (2,pen(x),x(7x+j)/2) (j = 3,5), (k,x^2,pen(x)) (k = 1,2,3,4,5,8,11).
(iii) Each natural number can be written as P(x,y,z) with x, y and z integers, where P(x,y,z) is either of the following polynomials: T(x)*x(5x+1)/2+T(y)+2*T(z), a*T(x)*pen(x)+pen(y)+pen(z) (a = 1,2,3,4), T(x)*pen(x)+pen(y)+3*pen(z), T(x)*pen(x)+pen(y)+4*pen(z), 2*T(x)*pen(x)+pen(y)+3*pen(z), pen(x)*x(5x+j)/2+pen(y)+3*pen(z) (j = 1,3), x(3x+2)*pen(x)+pen(y)+4*pen(z), pen(x)*x(7x+1)/2+pen(y)+pen(z), pen(x)*x(9x+7)/2+pen(y)+pen(z).
See also A270594 and A270705 for some other similar conjectures.

Examples

			a(1) = 1 since 1 = (-1)^2*T(-1) + 1^2 + T(0).
a(3) = 1 since 3 = 1^2*T(1) + 1^2 + T(1).
a(90) = 1 since 90 = 3^2*T(3) + 6^2 + T(0).
a(438) = 1 since 438 = 4^2*T(4) + 5^2 + T(22).
a(480) = 1 since 480 = 1^2*T(1) + 17^2 + T(19).
a(7108) = 1 since 1^2*T(1) + 69^2 + T(68).
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]]
    Do[r=0;Do[If[x!=0&&TQ[n-y^2-x^3*(x+1)/2],r=r+1],{y,1,Sqrt[n]},{x,-1-Floor[(2(n-y^2))^(1/4)],(2(n-y^2))^(1/4)}];Print[n," ",r];Continue,{n,1,90}]
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