A000797 -id:A000797 - cp's OEIS frontend

A104246 Minimal number of tetrahedral numbers (A000292(k) = k(k+1)(k+2)/6) needed to sum to n.

1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 2, 3, 4, 3, 3, 2, 3, 4, 4, 3, 3, 4, 5, 4, 4, 2, 1, 2, 3, 3, 2, 3, 4, 4, 3, 3, 2, 3, 4, 4, 2, 3, 4, 5, 3, 3, 2, 3, 4, 4, 3, 4, 5, 5, 1, 2, 3, 4, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 3, 4, 4, 3, 4

Offset: 1

Eric W. Weisstein, Feb 26 2005

nonn

According to Dickson, Pollock conjectures that a(n) <= 5 for all n. Watson shows that a(n) <= 8 for all n, and Salzer and Levine show that a(n) <= 5 for n <= 452479659. - N. J. A. Sloane, Jul 15 2011
Possible correction of the first comment by Sloane 2011: it appears to me from the linked reference by Salzer and Levine 1968 that 452479659 is instead the upper limit for sums of five Qx = Tx + x, where Tx are the tetrahedral numbers we want. They also mention an upper limit for sums of five Tx, which is: a(n) <= 5 for n <= 276976383. - Ewoud Dronkert, May 30 2024
If we use the greedy algorithm for this, we get A281367. - N. J. A. Sloane, Jan 30 2017
Could be extended with a(0) = 0, in analogy to A061336. Kim (2003, first row of table "d = 3" on p. 73) gives max {a(n)} = 5 as a "numerical result", but the value has no "* denoting exact values" (see Remark at end of paper), which means this could be incorrect. - M. F. Hasler, Mar 06 2017, edited Sep 22 2022

Dickson, L. E., History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 1952, see p. 13.

Lars Blomberg, Table of n, a(n) for n = 1..10000
Hyun Kwang Kim, On regular polytope numbers, Proc. Amer. Math. Soc. 131 (2003), pp. 65-75.
F. Pollock, On the Extension of the Principle of Fermat's Theorem of the Polygonal Numbers to the Higher Orders of Series Whose Ultimate Differences Are Constant. With a New Theorem Proposed, Applicable to All the Orders, Abs. Papers Commun. Roy. Soc. London 5, 922-924, 1843-1850.
H. E. Salzer and N. Levine, Table of Integers Not Exceeding 1000000 that are Not Expressible as the Sum of Four Tetrahedral Numbers, Math. Comp. 12, 141-144, 1958.
H. E. Salzer and N. Levine, Proof that every integer <= 452,479,659 is a sum of five numbers of the form Q_x = (x^3+5x)/6, x >= 0, Math. Comp., (1968), 191-192.
N. J. A. Sloane, Transforms.
G. L. Watson, Sums of eight values of a cubic polynomial, J. London Math. Soc., 27 (1952), 217-224.
Eric Weisstein's World of Mathematics, Tetrahedral Number.

Cf. A000292 (tetrahedral numbers, indices of 1s), A102795 (indices of 2s), A102796 (indices of 3s), A102797 (indices of 4s), A000797 (numbers that need 5 tetrahedral numbers).
See also A102798-A102806, A102855-A102858, A193101, A193105, A281367 (the "triangular nachos" numbers).
Cf. A061336 (analog for triangular numbers).

tet:=[seq((n^3-n)/6,n=1..20)];
LAGRANGE(tet,8, 120); # the LAGRANGE transform of a sequence is defined in A193101. - N. J. A. Sloane, Jul 15 2011
# alternative
N := 10000:
L := [seq(0,i=1..N)] :
# put 1's where tetrahedral numbers reside
for i from 1 to N do
    Aj := A000292(i) ;
    if Aj <= N then
        L := subsop(Aj=1,L) ;
    end if;
end do:
for a from 1 do
    # select positions of a's, skip forward by all available Aj and
    # if that addresses a not-yet-set position in the array put a+1 there.
    for i from 1 to N do
        if op(i,L) =a then
            for j from 1 do
                Aj := A000292(j) ;
                if i+Aj <=N and op(i+Aj,L) = 0 then
                    L := subsop(i+Aj=a+1,L) ;
                end if;
                if i +Aj > N then
                    break ;
                end if;
            end do:
        end if;
    end do:
    # if all L[] are non-zero, terminate the loop
    allset := true;
    for i from 1 to N do
        if op(i,L) = 0 then
            allset := false ;
            break ;
        end if;
    end do:
    if allset then
        break ;
    end if;
end do:
seq( L[i],i=1..N) ; # R. J. Mathar, Jun 06 2025

\\ available on request. - M. F. Hasler, Mar 06 2017

seq(N) = {
  my(a = vector(N, k, 8), T = k->(k*(k+1)*(k+2))\6);
  for (n = 1, N,
    my (k1 = sqrtnint((6*n)\8, 3), k2 = sqrtnint(6*n, 3));
    while(n < T(k2), k2--); if (n == T(k2), a[n] = 1; next());
    for (k = k1, k2, a[n] = min(a[n], a[n - T(k)] + 1))); a;
};
seq(102)  \\ Gheorghe Coserea, Mar 14 2017

Edited by N. J. A. Sloane, Jul 15 2011

Edited by M. F. Hasler, Mar 06 2017

A306460 Number of ways to write n as x(2x-1) + y(y+1)/2 + z(z+1)(z+2)/6, where x,y,z are nonnegative integers with x > 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 3, 3, 1, 2, 4, 3, 2, 1, 2, 6, 5, 1, 3, 3, 4, 6, 1, 1, 4, 6, 3, 3, 5, 3, 6, 4, 2, 3, 5, 5, 4, 6, 4, 2, 5, 5, 3, 5, 2, 6, 7, 3, 5, 5, 7, 5, 4, 2, 5, 8, 5, 3, 2, 6, 6, 4, 5, 5, 6, 7, 5, 5, 4, 6, 9, 6, 6, 5, 1, 7, 8, 3, 2, 6, 7, 5, 6, 5, 7, 8, 5, 3, 2, 6, 10, 6, 8, 7, 7, 5, 4, 6, 5, 5

Offset: 1

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture 1: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as the sum of a positive hexagonal number, a triangular number and a tetrahedral number.
We have verified a(n) > 0 for all n = 1..10^7.
Conjecture 2: Let c be 1 or 3. Then each n = 0,1,... can be written as c*x(x+1) + y*(y+1)/2 + z*(z+1)*(z+2)/6 with x,y,z nonnegative integers.
Conjecture 3: Let t(x) = x*(x+1)*(x+2)/6. Then each n = 0,1,... can be written as 2*t(w) + t(x) + t(y) + t(z) with w,x,y,z nonnegative integers.
We have verified Conjecture 3 for all n = 0..2*10^5. Clearly, Conjecture 3 implies Pollock's conjecture which states that any natural number is the sum of five tetrahedral numbers.

			a(3) = 1 with 3 = 1*(2*1-1) + 1*2/2 + 1*2*3/6.
a(14) = 1 with 14 = 1*(2*1-1) + 2*3/2 + 3*4*5/6.
a(75) = 1 with 75 = 5*(2*5-1) + 4*5/2 + 4*5*6/2.
a(349) = 1 with 349 = 5*(2*5-1) + 24*25/2 + 2*3*4/6.
a(369) = 1 with 369 = 4*(2*4-1) + 10*11/2 + 11*12*13/6.
a(495) = 1 with 495 = 8*(2*8-1) + 20*21/2 + 9*10*11/6.
a(642) = 1 with 642 = 16*(2*16-1) + 16*17/2 + 3*4*5/6.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Tetrahedral Number

Cf. A000217, A000292, A000384, A000797, A104246, A262813, A306462, A306471.

f[n_]:=f[n]=n(n+1)(n+2)/6;
TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[f[z]>=n, Goto[aa]]; Do[If[TQ[n-f[z]-x(2x-1)],r=r+1],{x,1,(Sqrt[8(n-f[z])+1]+1)/4}];Label[aa],{z,0,n}];tab=Append[tab,r],{n,1,100}];Print[tab]

A282172 Expansion of (Sum_{k>=0} x^(k(k+1)(k+2)/6))^5.

Original entry on oeis.org

1, 5, 10, 10, 10, 21, 30, 20, 15, 30, 35, 30, 40, 40, 35, 60, 65, 25, 30, 60, 46, 50, 80, 50, 55, 120, 95, 20, 60, 90, 60, 80, 100, 40, 80, 145, 85, 30, 90, 85, 105, 155, 100, 40, 155, 170, 90, 80, 100, 90, 171, 145, 40, 60, 140, 110, 125, 130, 80, 140, 250, 170, 70, 110, 140, 160, 190, 140, 90, 180, 220, 170, 95, 70, 110, 215

Offset: 0

Ilya Gutkovskiy, Feb 07 2017

nonn

Number of ways to write n as an ordered sum of 5 tetrahedral (or triangular pyramidal) numbers (A000292).

a(n) > 0 for all n ("Pollock's Conjecture").

			a(4) = 10 because we have:
[4, 0, 0, 0, 0]
[0, 4, 0, 0, 0]
[0, 0, 4, 0, 0]
[0, 0, 0, 4, 0]
[0, 0, 0, 0, 4]
[1, 1, 1, 1, 0]
[1, 1, 1, 0, 1]
[1, 1, 0, 1, 1]
[1, 0, 1, 1, 1]
[0, 1, 1, 1, 1]

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Ilya Gutkovskiy, Extended graphical example
Eric Weisstein's World of Mathematics, Pollock's Conjecture
Eric Weisstein's World of Mathematics, Tetrahedral Number
Index to sequences related to pyramidal numbers

Cf. A000292, A000797, A008439, A104246.

nmax = 75; CoefficientList[Series[(Sum[x^(k (k + 1) (k + 2)/6), {k, 0, nmax}])^5, {x, 0, nmax}], x]

G.f.: (Sum_{k>=0} x^(k*(k+1)*(k+2)/6))^5.

A306462 Number of ways to write n as C(2w,2) + C(x+2,3) + C(y+3,4) + C(z+4,5), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!), w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 3, 1, 1, 4, 7, 6, 2, 2, 6, 8, 5, 1, 2, 9, 11, 5, 1, 4, 9, 12, 7, 2, 4, 10, 12, 7, 4, 6, 10, 11, 6, 5, 5, 10, 15, 8, 4, 7, 11, 14, 9, 4, 5, 11, 14, 6, 6, 10, 15, 12, 5, 7, 8, 11, 14, 7, 5, 6, 11, 14, 12, 11, 6, 11, 15, 12, 7, 9, 18, 21, 12, 5, 5, 15, 19, 11, 3, 5

Offset: 1

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 14, 19.
We have verified a(n) > 0 for all n = 1..5*10^6.
See also A306471 and A306477 for similar conjectures.

			a(1) = 1 with 1 = C(2,2) + C(2,3) + C(3,4) + C(4,5).
a(4) = 1 with 4 = C(2,2) + C(3,3) + C(4,4) + C(5,5).
a(5) = 1 with 5 = C(2,2) + C(4,3) + C(3,4) + C(4,5).
a(14) = 1 with 14 = C(4,2) + C(3,3) + C(4,4) + C(6,5).
a(19) = 1 with 19 = C(6,2) + C(4,3) + C(3,4) + C(4,5).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000292, A000332, A000384, A000389, A000797, A262813, A306460, A306471, A306477.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==3;
tab={};Do[r=0;Do[If[f[5,z]>=n,Goto[cc]];Do[If[f[4,y]>=n-f[5,z],Goto[bb]];Do[If[f[3,x]>=n-f[5,z]-f[4,y],Goto[aa]];If[HQ[n-f[5,z]-f[4,y]-f[3,x]],r=r+1],{x,0,n-1-f[5,z]-f[4,y]}];Label[aa],{y,0,n-1-f[5,z]}];Label[bb],{z,0,n-1}];Label[cc];tab=Append[tab,r],{n,1,80}];Print[tab]

A306471 Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Original entry on oeis.org

1, 3, 3, 2, 4, 6, 5, 4, 4, 5, 7, 8, 6, 4, 5, 8, 8, 5, 4, 6, 7, 10, 10, 6, 6, 12, 13, 8, 7, 7, 6, 11, 9, 4, 3, 8, 16, 12, 8, 9, 9, 13, 14, 10, 7, 9, 18, 12, 6, 5, 4, 11, 10, 4, 2, 5, 19, 21, 11, 9, 13, 20, 16, 9, 6, 8, 17, 17, 4, 2, 9, 20, 17, 6, 9, 9, 15, 23, 14, 9, 15

Offset: 0

Zhi-Wei Sun, Feb 17 2019

nonn

Conjecture 1: a(n) > 1 for all n > 0.
We have verified a(n) > 0 for all n = 0..5*10^6.
Conjecture 2: For each r = 0, 1, any positive integer can be written as w^2 + C(x,3) + C(y,4) + C(z,5), where w,x,y,z are nonnegative integers with w - r even.
See also A306462 and A306477 for similar conjectures.

			a(0) = 1 with 0 = C(1,2) + C(2,3) + C(3,4) + C(4,5).
a(3) = 2 with 3 = C(3,2) + C(2,3) + C(3,4) + C(4,5) = C(1,2) + C(3,3) + C(4,4) + C(5,5).
a(54) = 2 with 54 = C(3,2) + C(7,3) + C(6,4) + C(5,5) = C(3,2) + C(5,3) + C(7,4) + C(6,5).
a(69) = 1 with 69 = C(3,2) + C(5,3) + C(7,4) + C(7,5) = C(3,2) + C(5,3) + C(3,4) + C(8,5).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000217, A000290, A000292, A000332, A000389, A000797, A014105, A262813, A306460, A306462, A306477.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;
tab={};Do[r=0;Do[If[f[5,z]>n,Goto[cc]];Do[If[f[4,y]>n-f[5,z],Goto[bb]];Do[If[f[3,x]>n-f[5,z]-f[4,y],Goto[aa]];If[HQ[n-f[5,z]-f[4,y]-f[3,x]],r=r+1],{x,0,n-f[5,z]-f[4,y]}];Label[aa],{y,0,n-f[5,z]}];Label[bb],{z,0,n}];Label[cc];tab=Append[tab,r],{n,0,80}];Print[tab]

A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 1, 2, 3, 3, 3, 4, 3, 2, 2, 2, 1, 2, 2, 4, 4, 4, 2, 2, 3, 2, 1, 4, 4, 4, 4, 4, 2, 1, 3, 4, 3, 4, 4, 4, 5, 3, 2, 3, 4, 2, 4, 5, 3, 2, 4, 2, 1, 1, 3, 4, 6, 4, 2, 3, 4, 2, 3, 5, 4, 5, 7, 5, 2, 4, 4, 4, 3, 3, 4, 6, 4, 4, 2, 2, 2, 4, 3, 6, 6, 5, 4, 6, 3, 2, 3, 6, 4, 6, 4, 4, 4, 4, 3, 3

Offset: 0

Zhi-Wei Sun, Feb 20 2019

nonn

Conjecture: a(n) > 0 for all n >= 0. In other words, each nonnegative integer can be written as the sum of a nonnegative cube and three tetrahedral numbers.
It seems that a(n) = 1 only for n = 0, 7, 17, 27, 34, 53, 54, 110, 118, 163, 207, 263, 270, 309, 362, 443, 1174, 1284.
We have verified a(n) > 0 for all n = 0..2*10^6.

			a(0) = 1 with 0 = 0^3 + C(2,3) + C(2,3) + C(2,3).
a(17) = 1 with 17 = 2^3 + C(3,3) + C(4,3) + C(4,3).
a(27) = 1 with 27 = 3^3 + C(2,3) + C(2,3) + C(2,3).
a(362) = 1 with 362 = 0^3 + C(6,3) + C(8,3) + C(13,3).
a(443) = 1 with 443 = 3^3 + C(5,3) + C(10,3) + C(13,3).
a(1174) = 1 with 1174 = 1^3 + C(9,3) + C(10,3) + C(19,3).
a(1284) = 1 with 1284 = 10^3 + C(7,3) + C(9,3) + C(11,3).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000292, A000578, A000797, A262813, A306460, A306462, A306471, A306477.

f[n_]:=f[n]=Binomial[n+2,3];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
tab={};Do[r=0;Do[If[f[x]>n/3,Goto[cc]];Do[If[f[y]>(n-f[x])/2,Goto[bb]];Do[If[f[z]>n-f[x]-f[y],Goto[aa]];If[CQ[n-f[x]-f[y]-f[z]],r=r+1],{z,y,n-f[x]-f[y]}];Label[aa],{y,x,(n-f[x])/2}];Label[bb],{x,0,n/3}];Label[cc];tab=Append[tab,r],{n,0,100}];Print[tab]

A102800 Let f(n) = A104246(n) be the minimal number of nonzero tetrahedral numbers that add to n; sequence gives numbers n for which f(n) <= 4.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 74

Offset: 1

Jud McCranie, Feb 26 2005

nonn

Cf. A000292, A000797, A104246, A102795, etc.

A356037 Conjecturally, a(n) is the smallest number m such that every natural number is a sum of at most m n-simplex numbers.

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 15, 15, 19, 24

Offset: 1

Mohammed Yaseen, Jul 24 2022

n-simplex numbers are {binomial(k,n); k>=n}.
This problem is the simplex number analog of Waring's problem.
a(2) = 3 was proposed by Fermat and proved by Gauss, see A061336.
Pollock conjectures that a(3) = 5. Salzer and Levine prove this for numbers up to 452479659. See A104246 and A000797.
Kim gives a(4)=8, a(5)=10, a(6)=13 and a(7)=15 (not proved).

			2-simplex numbers are {binomial(k,2); k>=2} = {1,3,6,10,...}, the triangular numbers. 3 is the smallest number m such that every natural number is a sum of at most m triangular numbers. So a(2)=3.
3-simplex numbers are {binomial(k,3); k>=3} = {1,4,10,20,...}, the tetrahedral numbers. 5 is presumed to be the smallest number m such that every natural number is a sum of at most m tetrahedral numbers. So a(3)=5.

Hyun Kwang Kim, On regular polytope numbers, Proc. Amer. Math. Soc. 131 (2003), p. 65-75.

Cf. A002804, A079611.
Minimal number of x-simplex numbers whose sum equals n: A061336 (x=2), A104246 (x=3), A283365 (x=4), A283370 (x=5).
x-simplex numbers: A000217 (x=2), A000292 (x=3), A000332 (x=4), A000389 (x=5), A000579 (x=6), A000580 (x=7), A000581 (x=8), A000582 (x=9).

cp's OEIS Frontend

A104246 Minimal number of tetrahedral numbers (A000292(k) = k(k+1)(k+2)/6) needed to sum to n.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

PARI

PARI

Extensions

A306460 Number of ways to write n as x*(2x-1) + y*(y+1)/2 + z*(z+1)*(z+2)/6, where x,y,z are nonnegative integers with x > 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A282172 Expansion of (Sum_{k>=0} x^(k*(k+1)*(k+2)/6))^5.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A306462 Number of ways to write n as C(2w,2) + C(x+2,3) + C(y+3,4) + C(z+4,5), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!), w is a positive integer and x,y,z are nonnegative integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A306471 Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A102800 Let f(n) = A104246(n) be the minimal number of nonzero tetrahedral numbers that add to n; sequence gives numbers n for which f(n) <= 4.

Views

Author

Keywords

Crossrefs

A356037 Conjecturally, a(n) is the smallest number m such that every natural number is a sum of at most m n-simplex numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A306460 Number of ways to write n as x(2x-1) + y(y+1)/2 + z(z+1)(z+2)/6, where x,y,z are nonnegative integers with x > 0.

A282172 Expansion of (Sum_{k>=0} x^(k(k+1)(k+2)/6))^5.