A133388 -id:A133388 - cp's OEIS frontend

A356208 a(n) is the number of occurrences of n in A133388.

2, 3, 4, 4, 5, 7, 6, 8, 8, 9, 9, 10, 10, 12, 13, 12, 12, 15, 14, 17, 16, 16, 17, 18, 19, 18, 19, 20, 18, 24, 20, 22, 25, 22, 27, 26, 23, 25, 25, 29, 26, 30, 27, 31, 32, 32, 24, 33, 33, 34, 32, 32, 35, 37, 36, 37, 38, 32, 35, 44, 36, 41, 41, 40, 42, 45, 39, 43, 42

Offset: 1

Hugo Pfoertner, Sep 07 2022

nonn

Cf. A000161, A001481, A133388, A356209.

from sympy.solvers.diophantine.diophantine import diop_DN
def A356208(n): return sum(1 for m in range(1,(n**2<<1)+1) if n==max((a for a, b in diop_DN(-1,m)),default=0)) # Chai Wah Wu, Sep 08 2022

A356209 a(n) is the position of the latest occurrence of n in A133388.

Original entry on oeis.org

2, 8, 18, 32, 41, 72, 98, 128, 162, 181, 242, 288, 313, 392, 421, 512, 514, 648, 722, 761, 882, 968, 1058, 1152, 1201, 1301, 1458, 1568, 1466, 1741, 1922, 2048, 2178, 2056, 2381, 2592, 2594, 2888, 2817, 3121, 3202, 3528, 3698, 3872, 3789, 4232, 4418, 4608, 4802, 4804, 5101

Offset: 1

Hugo Pfoertner, Sep 07 2022

nonn

Cf. A000161, A001481, A009003, A356208.

from sympy.solvers.diophantine.diophantine import diop_DN
def A356209(n):
    for m in range(n**2<<1,0,-1):
        if n==max((a for a, b in diop_DN(-1,m)),default=0):
            return m # Chai Wah Wu, Sep 08 2022

a(k) = 2*k^2 for k not in A009003.

A000161 Number of partitions of n into 2 squares.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 0, 0, 1, 0, 1, 0

Offset: 0

N. J. A. Sloane

Number of ways of writing n as a sum of 2 (possibly zero) squares when order does not matter.
Number of similar sublattices of square lattice with index n.
Let Pk = the number of partitions of n into k nonzero squares. Then we have A000161 = P0 + P1 + P2, A002635 = P0 + P1 + P2 + P3 + P4, A010052 = P1, A025426 = P2, A025427 = P3, A025428 = P4. - Charles R Greathouse IV, Mar 08 2010, amended by M. F. Hasler, Jan 25 2013
a(A022544(n))=0; a(A001481(n))>0; a(A125022(n))=1; a(A118882(n))>1. - Reinhard Zumkeller, Aug 16 2011

			25 = 3^2+4^2 = 5^2, so a(25) = 2.

J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 339

T. D. Noe, Table of n, a(n) for n = 0..10000
B. K. Agarwala and F. C. Auluck, Statistical mechanics and partitions into non-integral powers of integers, Proc. Camb. Phil. Soc., 47 (1951), 207-216. [Annotated scanned copy]
Henry Bottomley, Illustration of initial terms
R. T. Bumby, Sums of four squares, in Number theory (New York, 1991-1995), 1-8, Springer, New York, 1996.
J. H. Conway, E. M. Rains and N. J. A. Sloane, On the existence of similar sublattices, Canad. J. Math. 51 (1999), 1300-1306 (Abstract, pdf, ps).
Michael Gilleland, Some Self-Similar Integer Sequences
E. Grosswald, Representations of Integers as Sums of Squares, Springer-Verlag, NY, 1985, p. 84.
M. D. Hirschhorn, Some formulas for partitions into squares, Discrete Math, 211 (2000), pp. 225-228. [From _Ant King_, Oct 05 2010]
Index entries for sequences related to sublattices
Index entries for sequences related to sums of squares
Index entries for "core" sequences

Cf. A002654, A001481, A025427, A025428, A063725, A025426, A000290.
Cf. A000925, A247367.
Equivalent sequences for other numbers of squares: A010052 (1), A000164 (3), A002635 (4), A000174 (5).

a000161 n =
   sum $ map (a010052 . (n -)) $ takeWhile (<= n `div` 2) a000290_list
a000161_list = map a000161 [0..]
-- Reinhard Zumkeller, Aug 16 2011

A000161 := proc(n) local i,j,ans; ans := 0; for i from 0 to n do for j from i to n do if i^2+j^2=n then ans := ans+1 fi od od; RETURN(ans); end; [ seq(A000161(i), i=0..50) ];
A000161 := n -> nops( numtheory[sum2sqr](n) ); # M. F. Hasler, Nov 23 2007

Length[PowersRepresentations[ #,2,2]] &/@Range[0,150] (* Ant King, Oct 05 2010 *)

a(n)=sum(i=0,n,sum(j=0,i,if(i^2+j^2-n,0,1))) \\ for illustrative purpose

A000161(n)=sum(k=sqrtint((n-1)\2)+1,sqrtint(n),issquare(n-k^2)) \\ Charles R Greathouse IV, Mar 21 2014, improves earlier code by M. F. Hasler, Nov 23 2007

A000161(n)=#sum2sqr(n) \\ See A133388 for sum2sqr(). - M. F. Hasler, May 13 2018

from math import prod
from sympy import factorint
def A000161(n):
    f = factorint(n)
    return int(not any(e&1 for e in f.values())) + (((m:=prod(1 if p==2 else (e+1 if p&3==1 else (e+1)&1) for p, e in f.items()))+((((~n & n-1).bit_length()&1)<<1)-1 if m&1 else 0))>>1) if n else 1 # Chai Wah Wu, Sep 08 2022

a(n) = card { { a,b } c N | a^2+b^2 = n }. - M. F. Hasler, Nov 23 2007

Let f(n)= the number of divisors of n that are congruent to 1 modulo 4 minus the number of its divisors that are congruent to 3 modulo 4, and define delta(n) to be 1 if n is a perfect square and 0 otherwise. Then a(n)=1/2 (f(n)+delta(n)+delta(1/2 n)). - Ant King, Oct 05 2010

A136118 Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number.

Original entry on oeis.org

5, 4, 7, 12, 19, 17, 25, 20, 10, 28, 45, 42, 39, 17, 37, 21, 36, 35, 13, 33, 65, 28, 67, 32, 52, 40, 74, 31, 70, 85, 35, 16, 60, 70, 77, 68, 42, 30, 105, 76, 59, 26, 74, 49, 115, 19, 125, 115, 102, 110, 92, 56, 103, 29, 145, 100, 114, 77, 92, 47, 63, 108, 152, 95, 22, 116

Offset: 1

M. F. Hasler, Dec 25 2007

nonn

			a(1)=5 is the least integer m>0 such that A136117(1)-P(m) is a pentagonal number, namely P(7)-P(5)=70-35=35=P(5).
a(2)=4 is the least integer m>0 such that A136117(2)-P(m) is a pentagonal number, namely P(8)-P(4)=92-22=70=P(7).

Cf. A000326, A136112-A136117.

A136118vect(n,i=-1)=vector(n,k,until(0,for(j=2,#n=sum2sqr((i+=6)^2+1),n[j]%6==[5,5]||next;n=n[j];break(2)));n[1]\6+1) /* This uses sum2sqr(), cf. A133388. Below some simpler but much slower code. */
my(P=A000326(n)=n*(3*n-1)/2,isPent(t)=P(sqrtint(t*2\3)+1)==t); for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j))&print1(j",")||next(2)))

A303374 Numbers of the form a^4 + b^6, with integers a, b > 0.

Original entry on oeis.org

2, 17, 65, 80, 82, 145, 257, 320, 626, 689, 730, 745, 810, 985, 1297, 1354, 1360, 2025, 2402, 2465, 3130, 4097, 4112, 4160, 4177, 4352, 4721, 4825, 5392, 6497, 6562, 6625, 7290, 8192, 10001, 10064, 10657, 10729, 14096, 14642, 14705, 15370, 15626, 15641, 15706, 15881

Offset: 1

M. F. Hasler, Apr 22 2018

A subsequence of A000404 (a^2 + b^2), A055394 (a^2 + b^3), A111925 (a^4 + b^2), A100291 (a^4 + b^3), A303372 (a^2 + b^6).

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form. Maybe the most efficient way is to consider decompositions of n into sums of two positive squares (see sum2sqr in A133388), and check if one of the terms is a third power and the other a fourth power.

Cf. A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4).

Cf. A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303375 (a^5 + b^6).

Take[Flatten[Table[a^4+b^6,{a,20},{b,20}]]//Union,50] (* Harvey P. Dale, Jul 17 2025 *)

is(n,k=4,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,k)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
is(n,L=sum2sqr(n))={for(i=1,#L,L[i][1]&&for(j=1,2,ispower(L[i][j],3)&&issquare(L[i][3-j])&&return(L[i][j])))} \\ See A133388 for sum2sqr(). Much faster than the above for n >> 10^30.
A303374(L=10^5,k=4,m=6,S=[])={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k),S=setunion(S,[a^m+b^k])));S}

A136117 Pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

Original entry on oeis.org

70, 92, 852, 925, 1247, 1426, 1926, 2625, 3577, 5192, 6305, 6501, 7107, 7740, 7957, 8177, 8626, 9560, 10292, 12927, 13207, 14652, 15555, 16172, 18095, 20475, 20827, 21901, 22265, 22632, 23002, 23751, 24130, 28497, 29330, 31032, 33227, 33675

Offset: 1

M. F. Hasler, Dec 15 2007; corrected Dec 25 2007

nonn

It is conjectured that every integer and hence every pentagonal number, greater than 33066, hence greater than A000326(149) = 33227, can be represented as the sum of three pentagonal numbers. - Jonathan Vos Post, Dec 18 2007

			a(1)=70=P(7) is the least pentagonal number which can be written as sum of two other pentagonal numbers, P(7)=P(5)+P(5).

Cf. A000326, A136112-A136118, A007527.

P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */
isPent(t)=P(sqrtint(t<<1\3)+1)==t
for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j)) && print1(P(i)",") || next(2)))
/* The following is much faster, at the cost of implementing sum2sqr(), cf. A133388*/
A136117next(i)=i=sqrtint(i\3*2)*6+5; until(0, for(j=2,#t=sum2sqr((i+=6)^2+1), t[j]%6==[5,5] && break(2)));i^2\24
A136117vect(n,i)=vector(n,j,i=A136117next(i)) /* 2nd arg =0 by default but allows one to start elsewhere */
A136117(n,i)=until(!n--,i=A136117next(i));i \\ M. F. Hasler, Dec 25 2007

a(n) = A000326(A136116(n)) = A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.

A304433 Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

Original entry on oeis.org

5, 7, 8, 10, 12, 13, 14, 17, 20, 25, 26, 28, 29, 32, 33, 34, 37, 40, 41, 45, 48, 50, 52, 53, 56, 57, 58, 61, 63, 65, 68, 71, 72, 73, 74, 78, 80, 82, 85, 89, 90, 97, 98, 100, 101, 104, 105, 106, 109, 112, 113, 114, 116, 117, 122, 125, 126, 128

Offset: 1

M. F. Hasler, May 12 2018

nonn

Motivated by the search of solutions to a^n + b^(2n+2)/4 = (perfect square), which arises when searching solutions to x^n + y^(n+1) = z^(n+2) of the form x = a*z, y = b*z. It turns out that many solutions are of the form a^n = d (b^(n+1) + d), where d is a perfect power.

			5^3 = 125 = 4^2 + 11^2; 7^3 = 10^2 + 3^5; 8^3 = 13^2 + 7^3, ...

Robert Israel, Table of n, a(n) for n = 1..5970

Cf. A001597 (perfect powers), A076467 (cubes and higher powers), A304434, A304435, A304436 (analog for n^4, n^5, n^6).

N:= 200: # to get terms <= N
N3:= N^3:
P:= {seq(seq(x^k, k=3..floor(log[x](N3))), x=2..N)}:
filter:= proc(n) local n3, Pp, x, y;
  n3:= n^3;
  if remove(t -> subs(t, x)<=1 or subs(t, y)<=1 or subs(t, x-y)=0, [isolve(x^2+y^2=n3)]) <> [] then return true fi;
  Pp:= map(t ->n3-t, P minus {n3, n3/2});
   (Pp intersect P <> {}) or (select(issqr, Pp) <> {})
end proc:
select(filter, [$2..N]); # Robert Israel, Jun 01 2018

M = 200;
M3 = M^3;
P = Union@ Flatten@ Table[Table[x^k, {k, 3, Floor[Log[x, M3]]}], {x, 2, M}];
filterQ[n_] := Module[{n3, Pp, x, y}, n3 = n^3; If[Solve[x > 1 && y > 1 && x != y && x^2 + y^2 == n3, {x, y}, Integers] != {}, Return[True]]; Pp = n3 - (P ~Complement~ {n3, n3/2}); (Pp ~Intersection~ P) != {} || Select[ Pp, IntegerQ[Sqrt[#]]&] != {}];
Select[Range[2, M], filterQ] (* Jean-François Alcover, Jun 21 2020, after Robert Israel *)

L=200^3; P=List(); for(x=2, sqrtnint(L,3), for(k=3, logint(L, x), listput(P, x^k))); #P=Set(P) \\ This P = A076467 \ {1} = A111231 \ {0} up to limit L.
is(n,e=3)={for(i=1, #s=sum2sqr(n=n^e), vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=1, #P, n>P[i]||return; ispower(n-P[i])&& P[i]*2 != n && return(1))} \\ The above P must be computed up to L >= n^3. For sum2sqr() see A133388.

A304434 Numbers n such that n^4 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

Original entry on oeis.org

3, 5, 6, 9, 10, 12, 13, 14, 15, 17, 20, 24, 25, 26, 28, 29, 30, 34, 35, 36, 37, 39, 40, 41, 42, 45, 48, 50, 51, 52, 53, 55, 57, 58, 60, 61, 63, 65, 68, 70, 71, 72, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 96, 97, 98, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113

Offset: 1

M. F. Hasler, May 22 2018

nonn

Motivated by the search of solutions to a^n + b^(2n+2)/4 = (perfect square), which arises when searching solutions to x^n + y^(n+1) = z^(n+2) of the form x = a*z, y = b*z. It turns out that many solutions are of the form a^n = d (b^(n+1) + d), where d is a perfect power.

			3^4 = 2^5 + 7^2; 5^4 = 7^2 + 24^2, ...

Robert Israel, Table of n, a(n) for n = 1..1301

Cf. A304433, A001597 (perfect powers), A076467 (third or higher powers).

N:= 200: # to get terms <= N
N4:= N^4:
P:= {seq(seq(x^k,k=3..floor(log[x](N4))),x=2..floor(N4^(1/3)))}:
filter:= proc(n) local n4, Pp;
  n4:= n^4;
  if remove(t -> subs(t,x)<=1 or subs(t,y)<=1 or subs(t,x-y)=0, [isolve(x^2+y^2=n4)]) <> [] then return true fi;
  Pp:= map(t ->n4-t, P minus {n4, n4/2});
  (Pp intersect P <> {}) or (select(issqr,Pp) <> {})
end proc:
A:= select(filter, [$2..N]); # Robert Israel, May 24 2018

L=200^4; P=List(); for(x=2, sqrtnint(L,3), for(k=3, logint(L, x), listput(P, x^k))); #P=Set(P) \\ This P = A076467 \ {1} = A111231 \ {0} up to limit L.
is_A304434(n)={for(i=1, #s=sum2sqr(n=n^4), vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=1, #P, n>P[i]||return; ispower(n-P[i])&& P[i]*2 != n && return(1))} \\ The above P must be computed up to L >= n^4. For sum2sqr() see A133388.

A136116 Indices of pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

Original entry on oeis.org

7, 8, 24, 25, 29, 31, 36, 42, 49, 59, 65, 66, 69, 72, 73, 74, 76, 80, 83, 93, 94, 99, 102, 104, 110, 117, 118, 121, 122, 123, 124, 126, 127, 138, 140, 144, 149, 150, 152, 161, 163, 168, 169, 174, 175, 178, 181, 185, 188, 190, 195, 199, 203, 209, 210, 212, 213

Offset: 1

M. F. Hasler, Dec 15 2007; corrected Dec 25 2007

nonn

			a(1)=7 since P(7)=70 is the least pentagonal number which can be written as sum of two other pentagonal numbers, P(7)=P(5)+P(5).

Cf. A000326, A136112-A136118.

P(n)=n*(3*n-1)>>1 /* a.k.a. A000326 */
isPent(t)=P(sqrtint(t<<1\3)+1)==t
for(i=1,999,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j))&print1(i",") || next(2)))
/* The following are much faster, at the cost of implementing sum2sqr(), cf. A133388. */
A136116next(i)=i=6*i-1;until(0,for(j=2,#t=sum2sqr((i+=6)^2+1),t[j]%6==[5,5] && break(2))); i\6+1
A136116vect(n,i=0)=vector(n,j,i=A136116next(i))
A136116(n,i=0)=until(!n--,i=A136116next(i));i \\ M. F. Hasler, Dec 25 2007

A000326(a(n))=A000326(m)+A136114(m) where m is the index of the n-th nonzero term in A136114 or A136115.

A226539 Numbers which are the sum of two squared primes in exactly two ways (ignoring order).

Original entry on oeis.org

338, 410, 578, 650, 890, 1010, 1130, 1490, 1730, 1802, 1898, 1970, 2330, 2378, 2738, 3050, 3170, 3530, 3650, 3842, 3890, 4010, 4658, 4850, 5018, 5090, 5162, 5402, 5450, 5570, 5618, 5690, 5858, 6170, 6410, 6530, 6698, 7010, 7178, 7202, 7250, 7850, 7970, 8090

Offset: 1

Henk Koppelaar, Jun 10 2013

nonn

			338 = 7^2 + 17^2 = 13^2 + 13^2;
410 = 7^2 + 19^2 = 11^2 + 17^2.

Stan Wagon, Mathematica in Action, Springer, 2000 (2nd ed.), Ch. 17.5, pp. 375-378.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A054735 (restricted to twin primes), A037073, A069496.
Cf. A045636 (sum of two squared primes: a superset).
Cf. A214511 (least number having n representations).
Cf. A226562 (restricted to sums decomposed in exactly three ways).

Prime2PairsSum := p -> select(x ->`if`(andmap(isprime, x),true,false), numtheory:-sum2sqr(p)):
for n from 2 to 10^6 do
  if nops(Prime2PairsSum(n)) = 2 then print(n, Prime2PairsSum(n)) fi;
od;

Select[Range@10000, Length[Select[ PowersRepresentations[#, 2, 2], And @@ PrimeQ[#] &]] == 2 &] (* Giovanni Resta, Jun 11 2013 *)

select( is_A226539(n)={#[0|t<-sum2sqr(n),isprime(t[1])&&isprime(t[2])]==2}, [1..10^4]) \\ For more efficiency, apply selection to A045636. See A133388 for sum2sqr(). - M. F. Hasler, Dec 12 2019

a(25)-a(44) from Giovanni Resta, Jun 11 2013

cp's OEIS Frontend

A356208 a(n) is the number of occurrences of n in A133388.

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A356209 a(n) is the position of the latest occurrence of n in A133388.

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A000161 Number of partitions of n into 2 squares.

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A136118 Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number.

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A303374 Numbers of the form a^4 + b^6, with integers a, b > 0.

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A136117 Pentagonal numbers (A000326) which are the sum of 2 other positive pentagonal numbers.

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A304433 Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

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A304434 Numbers n such that n^4 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

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