A303637 -id:A303637 - cp's OEIS frontend

A303656 Number of ways to write n as a^2 + b^2 + 3^c + 5^d, where a,b,c,d are nonnegative integers with a <= b.

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

Offset: 1

Zhi-Wei Sun, Apr 27 2018

nonn

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two squares, a power of 3 and a power of 5.
It has been verified that a(n) > 0 for all n = 2..2*10^10.
It seems that any integer n > 1 also can be written as the sum of two squares, a power of 2 and a power of 3.
The author would like to offer 3500 US dollars as the prize for the first proof of his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Jun 05 2018
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 2.4*10^11. - Zhi-Wei Sun, Jul 30 2022

			a(2) = 1 with 2 = 0^2 + 0^2 + 3^0 + 5^0.
a(5) = 1 with 5 = 0^2 + 1^2 + 3^1 + 5^0.
a(25) = 1 with 25 = 1^2 + 4^2 + 3^1 + 5^1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000244, A000290, A000351, A001481, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303429, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303702, A303821.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;Do[If[QQ[n-3^k-5^m],Do[If[SQ[n-3^k-5^m-x^2],r=r+1],{x,0,Sqrt[(n-3^k-5^m)/2]}]],{k,0,Log[3,n]},{m,0,If[n==3^k,-1,Log[5,n-3^k]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303601 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + Bell(k) + Bell(m) with 0 <= a <= b and 0 < k <= m, where Bell(k) denotes the k-th Bell number A000110(k).

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 5, 5, 6, 4, 5, 7, 5, 4, 7, 7, 7, 8, 8, 5, 9, 10, 7, 6, 9, 8, 8, 6, 7, 10, 10, 9, 8, 7, 8, 9, 10, 6, 9, 11, 7, 6, 8, 9, 10, 7, 10, 8, 7, 8, 10, 10, 9, 10, 8, 9, 13, 14, 10, 11, 12, 12, 9, 9, 12, 11, 13, 11, 9

Offset: 1

Zhi-Wei Sun, Apr 26 2018

nonn

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be expressed as the sum of two triangular numbers and two Bell numbers.
This has been verified for all n = 2..7*10^8. Note that 111277 cannot be written as the sum of two squares and two Bell numbers.
As log(Bell(n)) is asymptotically equivalent to n*log(n), Bell numbers eventually grow faster than any exponential function.
See also A303389, A303540, A303543 and A303637 for similar conjectures.

			a(2) = 1 with 2 = 0*(0+1)/2 + 0*(0+1)/2 + Bell(1) + Bell(1).
a(3) = 2 with 3 =  0*(0+1)/2 + 1*(1+1)/2 + Bell(1) + Bell(1) = 0*(0+1)/2 + 0*(0+1)/2 + Bell(1) + Bell(2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Cf. A000110, A000217, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303637.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
b[n_]:=b[n]=BellB[n];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;k=1;Label[bb];If[b[k]>n,Goto[aa]];Do[If[QQ[4(n-b[k]-b[j])+1],Do[If[TQ[n-b[k]-b[j]-x(x+1)/2],r=r+1],{x,0,(Sqrt[4(n-b[k]-b[j])+1]-1)/2}]],{j,1,k}];k=k+1;Goto[bb];Label[aa];
tab=Append[tab,r],{n,1,70}];Print[tab]

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))5^m squarefree.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 2, 1, 3, 1, 2, 2, 2, 1, 3, 3, 3, 2, 4, 2, 3, 2, 5, 2, 4, 2, 3, 3, 3, 2, 4, 3, 5, 1, 7, 4, 4, 3, 7, 2, 4, 3, 8, 4, 7, 4, 6, 3, 7, 3, 6, 4, 5, 3, 5, 4, 5, 2, 7, 3, 5, 4, 8, 4, 5, 3, 5, 5, 8, 6, 6, 6, 9, 3, 9, 7, 6, 6, 8, 5, 6, 4, 6, 8, 7, 6, 8, 7, 4

Offset: 1

Zhi-Wei Sun, May 06 2018

nonn

Conjecture: a(n) > 0 for all n > 7.
This has been verified for n up to 2*10^10.
See also A303821, A303934, A303949, A304031 and A304122 for related information, and A304034 for a similar conjecture.
The author would like to offer 2500 US dollars as the prize to the first proof of the conjecture, and 250 US dollars as the prize to the first explicit counterexample. - Zhi-Wei Sun, May 08 2018

			a(6) = 1 since 6 = 3 + 2^1 + 5^0 with 3 an odd prime and 2^1 + 5^0 = 3 squarefree.
a(15) = 1 since 15 = 5 + 2^3 + 2*5^0 with 5 an odd prime and 2^3 + 2*5^0 = 2*5 squarefree.
a(35) = 1 since 35 = 29 + 2^2 + 2*5^0 with 29 an odd prime and 2^2 + 2*5^0 = 2*3 squarefree.
a(91) = 1 since 91 = 17 + 2^6 + 2*5^1 with 17 an odd prime and 2^6 + 2*5^1 = 2*37 squarefree.
a(9574899) = 1 since 9574899 = 9050609 + 2^19 + 2*5^0 with 9050609 an odd prime and 2^19 + 2*5^0 = 2*5*13*37*109 squarefree.
a(6447154629) = 2 since 6447154629 = 6447121859 + 2^15 + 2*5^0 with 6447121859 prime and 2^15 + 2*5^0 = 2*5*29*113 squarefree, and 6447154629 = 5958840611 + 2^15 + 2*5^12 with 5958840611 prime and 2^15 + 2*5^12 = 2*17*41*433*809 squarefree.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv, arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000351, A005117, A098983, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821, A303932, A303934, A303949, A304031, A304032, A304034, A304122.

PQ[n_]:=n>2&&PrimeQ[n];
tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*5^m]&&PQ[n-2^k-(1+Mod[n,2])*5^m],r=r+1],{k,0,Log[2,n]},{m,0,If[2^k==n,-1,Log[5,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))3^m squarefree.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 3, 1, 4, 2, 5, 1, 3, 2, 5, 1, 7, 3, 3, 4, 4, 4, 6, 2, 3, 5, 6, 2, 7, 3, 5, 5, 6, 5, 9, 3, 4, 6, 7, 2, 12, 2, 5, 6, 7, 4, 10, 3, 3, 5, 8, 2, 8, 3, 4, 6, 8, 5, 9, 4, 2, 7, 7, 3, 13, 5, 5, 9, 7, 5, 13, 3, 6, 10, 7, 5, 10, 5, 7, 7, 9, 8, 13

Offset: 1

Zhi-Wei Sun, May 06 2018

nonn

Conjecture: a(n) > 0 for all n > 11.
This has been verified for n up to 10^10.
See also A304081 for a similar conjecture.

			a(8) = 1 since 8 = 3 + 2^1 + 3^1 with 3 prime and 2^1 + 3^1 = 5 squarefree.
a(13) = 1 since 13 = 3 + 2^2 + 2*3^1 with 3 prime and 2^2 + 2*3^1 = 2*5 squarefree.
a(19) = 1 since 19 = 5 + 2^3 + 2*3^1 with 5 prime and 2^3 + 2*3^1 = 2*7 squarefree.
a(23) = 1 since 23 = 13 + 2^2 + 2*3^1 with 13 prime and 2^2 + 2*3 = 2*5 squarefree.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000224, A005117, A118955, A155216, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821, A303932, A303934, A303949, A304031, A304032, A304081.

tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*3^m]&&PrimeQ[n-2^k-(1+Mod[n,2])*3^m],r=r+1],{k,1,Log[2,n]},{m,1,If[2^k==n,-1,Log[3,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303702 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 5, 7, 6, 6, 9, 9, 5, 8, 9, 6, 9, 11, 8, 10, 11, 7, 12, 15, 8, 10, 12, 7, 10, 9, 8, 12, 11, 5, 12, 16, 7, 13, 17, 8, 10, 15, 10, 13, 14, 10, 12, 17, 7, 12, 18, 11, 13, 17, 10, 13, 20, 11, 14, 17, 8, 10, 16, 7, 10

Offset: 1

Zhi-Wei Sun, Apr 29 2018

nonn

Conjecture: a(n) > 0 for all n > 1. In other words, any even number greater than 2 can be written as the sum of a prime, a power of 2 and a power of 3.
It has been verified that a(n) > 0 for all n = 2..3*10^9.
a(n) > 0 for n <= 10^11. - Jud McCranie, Jun 25 2023
a(n) > 0 for n < 10^12. - Jud McCranie, Jul 11 2023
a(n) > 0 for n <= 4*10^12. - Jud McCranie, Aug 17 2023

			a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 2 prime.
a(3) = 2 since 2*3 = 2 + 2^0 + 3^1 = 3 + 2^1 + 3^0 with 2 and 3 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000244, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660.

tab={};Do[r=0;Do[If[PrimeQ[2n-2^x-3^y],r=r+1],{x,0,Log[2,2n-1]},{y,0,Log[3,2n-2^x]}];tab=Append[tab,r],{n,1,65}];Print[tab]

A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 4, 4, 5, 3, 6, 5, 5, 6, 6, 4, 7, 6, 7, 7, 10, 4, 9, 10, 6, 10, 8, 5, 8, 6, 7, 7, 9, 5, 8, 11, 6, 10, 11, 6, 11, 8, 6, 8, 11, 4, 9, 9, 7, 6, 11, 6, 7, 11, 7, 10, 11, 5, 11, 9, 6, 7, 6, 6, 5, 12, 7, 10, 15, 8, 15, 10, 11, 13, 11, 7, 9, 8, 9, 12, 14

Offset: 1

Zhi-Wei Sun, May 01 2018

nonn

Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 4, we can write 2*n as p + 2^x + 5^y, where p is an odd prime, and x and y are positive integers.
This has been verified for n up to 10^10.
See also A303934 and A304081 for further refinements, and A303932 and A304034 for similar conjectures.

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime.
a(5616) = 2 since 2*5616 = 9059 + 2^11 + 5^3 = 10979 + 2^7 + 5^3 with 9059 and 10979 both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000351, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303932, A303934, A304034, A304081.

tab={};Do[r=0;Do[If[PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303660 Number of ways to write 2*n+1 as p + 3^k + 5^m, where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 4, 4, 5, 3, 4, 4, 3, 6, 7, 5, 6, 8, 5, 5, 9, 6, 5, 8, 3, 6, 8, 4, 4, 7, 6, 4, 8, 6, 5, 9, 4, 4, 8, 3, 6, 8, 7, 4, 9, 6, 4, 9, 5, 5, 9, 6, 6, 11, 7, 7, 9, 5, 3, 8, 5, 3, 9, 7, 7, 11, 8, 8, 12

Offset: 1

Zhi-Wei Sun, Apr 28 2018

nonn

Note that a(21323543) = 0, i.e., the odd number 2*21323543 + 1 = 42647087 cannot be written as the sum of a prime, a power of 3 and a power of 5.

			a(2) = 1 since 2*2+1 = 3 + 3^0 + 5^0 with 3 prime.
a(3) = 2 since 2*3+1 = 3 + 3^1 + 5^0 = 5 + 3^0 + 5^0 with 3 and 5 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.

Cf. A000040, A000244, A000351, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656.

tab={};Do[r=0;Do[If[PrimeQ[2n+1-5^x-3^y],r=r+1],{x,0,Log[5,2n]},{y,0,Log[3,2n+1-5^x]}];tab=Append[tab,r],{n,1,70}];Print[tab]

A308566 Number of ways to write n as w^2 + x(x+1) + 4^y5^z with w,x,y,z nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jun 07 2019

nonn

Recall an observation of Euler: {w^2 + x*(x+1): w,x = 0,1,2,...} = {a*(a+1)/2 + b*(b+1)/2: a,b = 0,1,...}.
Conjecture: a(n) > 0 for all n > 0. Equivalently, each n = 1,2,3,... can be written as a*(a+1)/2 + b*(b+1)/2 + 4^c*5^d with a,b,c,d nonnegative integers.
See also A308584 for a similar conjecture.
We have verified a(n) > 0 for all n = 1..5*10^8.
a(n) > 0 for 0 < n < 10^10. - Giovanni Resta, Jun 08 2019

			a(1) = 1 with 1 = 0^2 + 0*1 + 4^0*5^0.
a(2) = 1 with 2 = 1^2 + 0*1 + 4^0*5^0.
a(3) = 1 with 3 = 0^2 + 1*2 + 4^0*5^0.
a(9) = 1 with 9 = 2^2 + 0*1 + 4^0*5^1.
a(303) = 1 with 303 = 16^2 + 6*7 + 4^0*5^1.
a(585) = 1 with 585 = 5^2 + 15*16 + 4^3*5^1.
a(37863) = 2 with 37863 = 166^2 + 101*102 + 4^0*5^1 = 179^2 + 26*27 + 4^5*5^1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000290, A000302, A000351, A002378, A303656, A303637, A308411, A308547, A308584.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-4^k*5^m-x(x+1)],r=r+1],{k,0,Log[4,n]},{m,0,Log[5,n/4^k]},{x,0,(Sqrt[4(n-4^k*5^m)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]

A308584 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + 5^c*8^d, where a,b,c,d are nonnegative integers with a <= b.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Jun 08 2019

nonn

Conjecture: a(n) > 0 for all n > 0. Equivalently, each n = 1,2,3,... can be written as w^2 + x*(x+1) + 5^y*8^z with w,x,y,z nonnegative integers.
We have verified a(n) > 0 for all n = 1..4*10^8.
See also A308566 for a similar conjecture.
a(n) > 0 for all 0 < n < 10^10. - Giovanni Resta, Jun 10 2019

			a(13) = 1 with 13 = 3*4/2 + 3*4/2 + 5^0*8^0.
a(48) = 1 with 48 = 5*6/2 + 7*8/2 + 5^1*8^0.
a(87) = 1 with 87 = 1*2/2 + 12*13/2 + 5^0*8^1.
a(90) = 1 with 90 = 4*5/2 + 10*11/2 + 5^2*8^0.
a(423) = 1 with 423 = 9*10/2 + 22*23/2 + 5^3*8^0.
a(517) = 1 with 517 = 17*18/2 + 24*25/2 + 5^0*8^2.
a(985) = 1 with 985 = 19*20/2 + 34*35/2 + 5^2*8^1.
a(2694) = 1 with 2694 = 7*8/2 + 68*69/2 + 5^1*8^2.
a(42507) = 1 with 42507 = 178*179/2 + 223*224/2 + 5^2*8^2.
a(544729) = 1 with 544729 = 551*552/2 + 857*858/2 + 5^5*8^1.
a(913870) = 1 with 913870 = 559*560/2 + 700*701/2 + 5^3*8^4.
a(1843782) = 1 with 1843782 = 808*809/2 + 1668*1669/2 + 5^6*8^1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Cf. A000217, A000351, A001018, A303656, A303637, A308411, A308547, A308566.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[TQ[n-5^k*8^m-x(x+1)/2],r=r+1],{k,0,Log[5,n]},{m,0,Log[8,n/5^k]},{x,0,(Sqrt[4(n-5^k*8^m)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]

A303932 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic residue modulo p, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 2, 3, 3, 1, 3, 5, 2, 1, 4, 2, 1, 4, 3, 4, 4, 2, 3, 7, 4, 2, 6, 3, 2, 4, 4, 3, 3, 2, 4, 6, 2, 1, 6, 2, 2, 6, 5, 6, 5, 5, 6, 8, 3, 5, 8, 5, 3, 7, 6, 5, 7, 6, 9, 7, 5, 7, 7, 3, 5, 9, 5, 7, 9, 6, 11, 10, 5, 11, 10, 4, 5, 13, 3, 5

Offset: 1

Zhi-Wei Sun, May 02 2018

nonn

Conjecture 1. a(n) > 0 for all n > 1, i.e., any even number greater than two can be written as the sum of a power 2, a power of 3 and a prime p with 11 a quadratic residue modulo p.
Conjecture 2. For any integer n > 2, we can write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic nonresidue modulo p, and k and m are nonnegative integers.
We have verified Conjectures 1 and 2 for n up to 5*10^8.

			a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 11 a quadratic residue modulo the prime 2.
a(3) = 1 since 2*3 = 2 + 2^0 + 3^1 with 11 a quadratic residue modulo the prime 2.
a(10) = 1 since 2*10 = 7 + 2^2 + 3^2 with 11 a quadratic residue modulo the prime 7.
a(14) = 1 since 2*14 = 19 + 2^3 + 3^0 with 11 a quadratic residue modulo the prime 19.
a(17) = 1 since 2*17 = 5 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 5.
a(38) = 1 since 2*38 = 37 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 37.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000244, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821.

PQ[n_]:=PQ[n]=n==2||(n>2&&PrimeQ[n]&&JacobiSymbol[11,n]==1);
tab={};Do[r=0;Do[If[PQ[2n-2^k-3^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[3,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

cp's OEIS Frontend

A303656 Number of ways to write n as a^2 + b^2 + 3^c + 5^d, where a,b,c,d are nonnegative integers with a <= b.

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A303601 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + Bell(k) + Bell(m) with 0 <= a <= b and 0 < k <= m, where Bell(k) denotes the k-th Bell number A000110(k).

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A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))*5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))*5^m squarefree.

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A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))*3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))*3^m squarefree.

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A303702 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime, and k and m are nonnegative integers.

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A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

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A303660 Number of ways to write 2*n+1 as p + 3^k + 5^m, where p is a prime, and k and m are nonnegative integers.

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A308566 Number of ways to write n as w^2 + x*(x+1) + 4^y*5^z with w,x,y,z nonnegative integers.

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A303601 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + Bell(k) + Bell(m) with 0 <= a <= b and 0 < k <= m, where Bell(k) denotes the k-th Bell number A000110(k).

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))5^m squarefree.

A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))3^m squarefree.

A308566 Number of ways to write n as w^2 + x(x+1) + 4^y5^z with w,x,y,z nonnegative integers.

A308584 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + 5^c*8^d, where a,b,c,d are nonnegative integers with a <= b.