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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A092362 Number of partitions of n^2 into squares greater than 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 3, 5, 8, 11, 28, 44, 94, 167, 354, 643, 1314, 2412, 4792, 8981, 17374, 32566, 62008, 115702, 217040, 402396, 745795, 1372266, 2517983, 4595652, 8354350, 15125316, 27265107, 48972467, 87584837, 156119631, 277152178, 490437445, 864534950
Offset: 0

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Author

Reinhard Zumkeller, Mar 19 2004

Keywords

Comments

a(n) = A078134(A000290(n)).

Examples

			a(6) = 5: 6^2 = 36 = 16+16+4 = 16+4+4+4+4+4 = 9+9+9+9 = 4+4+4+4+4+4+4+4+4.
		

Crossrefs

Programs

  • Maple
    b:=proc(n, i) option remember; `if`(n=0, 1,
         `if`(i<2, 0, b(n, i-1) +`if`(i^2>n, 0, b(n-i^2, i))))
       end:
    a:= n-> b(n^2, n):
    seq(a(n), n=0..50);  # Alois P. Heinz, Apr 15 2013
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<2, 0, b[n, i-1] + If[i^2>n, 0, b[n-i^2, i]]]]; a[n_] := b[n^2, n]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Nov 11 2015, after Alois P. Heinz *)

Formula

a(n) ~ exp(3*Pi^(1/3) * Zeta(3/2)^(2/3) * n^(2/3) / 2^(4/3)) * Zeta(3/2)^(4/3) / (2^(11/3) * sqrt(3) * Pi^(5/6) * n^(11/3)). - Vaclav Kotesovec, Apr 10 2017

Extensions

Corrected a(0) and more terms from Alois P. Heinz, Apr 15 2013

A111178 Number of partitions of n into positive numbers one less than a square.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 4, 1, 2, 4, 1, 2, 5, 2, 4, 5, 2, 5, 5, 2, 6, 7, 4, 6, 7, 5, 6, 8, 6, 8, 12, 6, 9, 13, 6, 10, 15, 8, 14, 15, 9, 16, 16, 10, 18, 21, 14, 19, 22, 16, 20, 24, 19, 25, 30, 20, 27, 33, 21, 29, 39, 26, 37, 40, 28, 42, 42, 31, 48
Offset: 0

Views

Author

Wouter Meeussen, Oct 22 2005

Keywords

Comments

Also limiting form of the number of representations of n into k positive squares for k decreasing from n to 1, or Table[Count[SumOfSquaresRepresentations[k,n], {a_,}/;a>0], {n,100,100}, {k,100,40,-1}]. (Franklin T. Adams-Watters: replacing k^2 ones by the value k^2 changes the count by k^2-1).
a(n) = A243148(2n,n). - Alois P. Heinz, May 30 2014

Crossrefs

Programs

  • Haskell
    a111178 = p $ tail a005563_list where
       p _          0 = 1
       p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
    -- Reinhard Zumkeller, Apr 02 2014
  • Maple
    b:= proc(n, i) option remember;
          `if`(n=0, 1, `if`(i<2, 0, b(n, i-1)+
          `if`(i^2>n+1, 0, b(n+1-i^2, i))))
        end:
    a:= n-> b(n, isqrt(n)):
    seq(a(n), n=0..100);  # Alois P. Heinz, May 30 2014
  • Mathematica
    nn = 100; CoefficientList[Series[Product[1/(1 - x^(k^2 - 1)), {k, 2, nn}], {x, 0, nn}], x] (* corrected by T. D. Noe, Feb 22 2012 *)
    b[n_, i_] := b[n, i] = If[n==0, 1, If[i<2, 0, b[n, i-1] + If[i^2>n+1, 0, b[n+1-i^2, i]]]]; a[n_] := b[n, Round[Sqrt[n]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 16 2017, after Alois P. Heinz *)

Formula

G.f.: Product_{k>=2} 1/(1-x^(k^2-1)).

A134755 Minimal number such that all greater numbers can be written as sums of squares of primes in more than n ways.

Original entry on oeis.org

23, 39, 55, 64, 68, 73, 80, 84, 91, 96, 100, 105, 109, 113, 114, 118, 122, 123, 127, 131, 132, 136, 140, 140, 144, 145, 145, 149, 149, 153, 154, 156, 158, 160, 163, 164, 167, 168, 168, 172, 172, 176, 176, 176, 180, 180, 181, 181, 185, 185, 185, 189, 189, 190
Offset: 0

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Author

Hieronymus Fischer, Nov 11 2007

Keywords

Comments

The sequence is well-defined, in that a(n) exists for all n>=0. Proof by induction: a(0) exists. We set b(j):=number of ways to write j as sum of squares of primes (=A090677). If a(n) exists, then b(j)>n for all j>a(n). Setting m:=a(n)+1, we find that there are n+1 sum of squares of primes B(0,i), 1<=i<=n+1, with m=B(0,i).
Further there are n+1 such sum expressions B(1,i), B(2,i) and B(3,i), 1<=i<=n+1, representing m+1, m+2 and m+3, respectively. For all j>a(n) we have j=m+4*floor((j-m)/4)+(j-m) mod 4. Thus j=m+r+s*2^2, where r=0,1,2 or 3. Hence n can be written B(r,i)+s*2^2 and there are n+1 such representations.
Let q be the maximal prime number (to be squared) occurring as a term within those sum expressions B(r,i), 0<=r<=3,1<=i<=n+1. We select a prime number p>q and we set c:=a(n)+p^2. For j>c, we have the n+1 representations B(r(j),i)+s(j)*2^2. Additionally, for j-p^2 (which is >a(n)) there are also n+1 representations B(r_p,i)+s_p*2^2, where r_p:=r(j-p^2), s_p:=s(j-p^2).
Thus j can be written B(r(j),i)+s(j)*2^2, 1<=i<=n+1 and B(r_p,i)+s_p*2^2+p^2, 1<=i<=n+1. By choice of p all these sum representations of j are different, which implies, that there are 2n+2 such representations. It follows b(j)>2n+2>n+1 for all j>c, which implies, that a(n+1) exists.

Examples

			a(0)=23, since numbers >23 can be written as sum of squares of primes.
a(1)=39, since there are at least two ways, to write a number >39 as a sum of squares of primes.
		

Crossrefs

Formula

a(n)=min( m | A090677(j)>n for all j>m).

A033183 a(n) = number of pairs (p,q) such that 4*p + 9*q = n.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2
Offset: 0

Views

Author

Michel Tixier (tixier(AT)dyadel.net)

Keywords

Comments

From Reinhard Zumkeller, Nov 07 2009: (Start)
In other words: number of partitions into 4 or 9;
a(n) <= A078134(n); a(A078135(n)) = 0;
a(A167632(n)) = n and a(m) < n for m < A167632(n). (End)

Crossrefs

Cf. A033182.

Programs

  • Mathematica
    CoefficientList[Series[1/((1-x^4)(1-x^9)),{x,0,80}],x] (* or  *) LinearRecurrence[{0,0,0,1,0,0,0,0,1,0,0,0,-1}, {1,0,0,0,1,0,0,0,1,1,0,0,1}, 80] (* Harvey P. Dale, Oct 13 2012 *)

Formula

a(n) = [ 7 n/9 ]+1+[ -3 n/4 ].
G.f.: 1/((1-x^4)*(1-x^9)). - Vladeta Jovovic, Nov 12 2004
a(n) = a(n-4) + a(n-9) - a(n-13). - R. J. Mathar, Dec 04 2011

A134754 Minimal number such that all greater numbers can be written as sums of squares >1 in more than n ways.

Original entry on oeis.org

23, 39, 39, 55, 55, 55, 59, 59, 63, 71, 71, 71, 71, 75, 75, 75, 75, 79, 79, 87, 87, 87, 87, 87, 91, 91, 91, 91, 91, 91, 95, 95, 95, 95, 95, 96, 96, 99, 99, 103, 103, 103, 103, 103, 103, 103, 107, 107, 107, 107, 107, 107, 107, 107, 111, 111, 111, 111, 111, 111, 111
Offset: 0

Views

Author

Hieronymus Fischer, Nov 11 2007

Keywords

Comments

The sequence is well-defined, in that a(n) exists for all n>=0. For the reasoning see A078134.

Examples

			a(0)=23, since numbers >23 can be written as sum of squares >1.
a(2)=39, since there are at least three ways, to write a number >39 as a sum of squares >1.
		

Crossrefs

Formula

a(n)=min( m | A078134(j)>n for all j>m).

A093115 Number of partitions of n^2 into squares not greater than n.

Original entry on oeis.org

1, 1, 1, 1, 5, 7, 10, 13, 17, 108, 159, 228, 317, 430, 572, 748, 5753, 8125, 11266, 15376, 20672, 27430, 35942, 46575, 59717, 523905, 708028, 946875, 1253880, 1645224, 2140099, 2761318, 3535658, 4494602, 5674753, 7118724, 69766770, 90940578, 117756370
Offset: 0

Views

Author

Reinhard Zumkeller, Mar 21 2004

Keywords

Examples

			n=6: 6^2 = 9*2^2 = 8*2^2+4*1^2 = 7*2^2+8*1^2 = 6*2^2+12*1^2 = 5*2^2+16*1^2 = 4*2^2+20*1^2 = 3*2^2+24*1^2 = 2*2^2+28*1^2 = 1*2^2+32*1^2 = 36*1^2, therefore a(6)=10.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1,
         `if`(i<1, 0, b(n, i-1) +`if`(i^2>n, 0, b(n-i^2, i))))
        end:
    a:= proc(n) local r; r:= isqrt(n);
          b(n^2, r-`if`(r^2>n, 1, 0))
        end:
    seq(a(n), n=0..50);  # Alois P. Heinz, Apr 15 2013
  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, b[n, i-1] + If[i^2 > n, 0, b[n-i^2, i]]]]; a[n_] := (r = Sqrt[n] // Floor; b[n^2, r - If[r^2 > n, 1, 0]]); Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jul 29 2015, after Alois P. Heinz *)

Formula

Coefficient of x^(n^2) in the series expansion of Product_{k=1..floor(sqrt(n))} 1/(1 - x^(k^2)). - Vladeta Jovovic, Mar 24 2004

Extensions

More terms from Vladeta Jovovic, Mar 24 2004
Corrected a(0) by Alois P. Heinz, Apr 15 2013

A093116 Number of partitions of n^2 into squares not less than n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 2, 5, 4, 4, 5, 9, 15, 23, 24, 13, 20, 32, 55, 84, 113, 185, 303, 545, 167, 298, 435, 716, 1055, 1701, 2584, 4213, 6471, 10218, 15884, 4856, 7376, 11231, 17221, 26054, 39583, 60109, 91622, 138569, 209951, 318368, 483098, 730183
Offset: 0

Views

Author

Reinhard Zumkeller, Mar 21 2004

Keywords

Examples

			n=10: 10^2 = 100 = 64+36 = 36+16+16+16+16 = 25+25+25+25, all other partitions of 100 into squares contain parts < 10, therefore a(10) = 4.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1,
         `if`(i^2>n, 0, b(n, i+1) +b(n-i^2, i)))
        end:
    a:= proc(n) local r; r:= isqrt(n);
          b(n^2, r+`if`(r^2Alois P. Heinz, Apr 15 2013
  • Mathematica
    b[n_, i_] := b[n, i] = If[n==0, 1, If[i^2>n, 0, b[n, i+1] + b[n-i^2, i]]]; a[n_] := With[{r = Sqrt[n]//Floor}, b[n^2, r + If[r^2Jean-François Alcover, Oct 26 2015, after Alois P. Heinz *)

A302833 Expansion of (1/(1 - x))*Product_{k>=1} 1/(1 - x^(k^2)).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 12, 15, 19, 23, 27, 32, 38, 44, 50, 58, 67, 77, 87, 99, 112, 126, 140, 156, 175, 195, 216, 239, 265, 292, 320, 351, 385, 422, 460, 503, 549, 598, 648, 703, 763, 826, 892, 963, 1041, 1122, 1206, 1296, 1394, 1498, 1605, 1721, 1845, 1977, 2112, 2256, 2410, 2573
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 13 2018

Keywords

Comments

Partial sums of A001156.
Number of partitions of n into squares if there are two kinds of 1's.

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, n+1,
          b(n, i-1)+ `if`(i^2>n, 0, b(n-i^2, i)))
        end:
    a:= n-> b(n, isqrt(n)):
    seq(a(n), n=0..100);  # Alois P. Heinz, Apr 13 2018
  • Mathematica
    nmax = 58; CoefficientList[Series[1/(1 - x) Product[1/(1 - x^k^2), {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 58; CoefficientList[Series[1/(1 - x) Sum[x^j^2/Product[(1 - x^k^2), {k, 1, j}], {j, 0, nmax}], {x, 0, nmax}], x]

Formula

G.f.: (1/(1 - x))*Sum_{j>=0} x^(j^2)/Product_{k=1..j} (1 - x^(k^2)).
From Vaclav Kotesovec, Apr 13 2018: (Start)
a(n) ~ exp(3*Pi^(1/3) * Zeta(3/2)^(2/3) * n^(1/3) / 2^(4/3)) / (2*Pi^(3/2) * sqrt(3*n)).
a(n) ~ 2^(4/3) * n^(2/3) / (Pi^(1/3) * Zeta(3/2)^(2/3)) * A001156(n). (End)

A078138 Primes which can be written as sum of squares > 1.

Original entry on oeis.org

13, 17, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 19 2002

Keywords

Comments

By Sylvester's solution to the Frobenius problem, all integers greater than 4*9 - 4 - 9 = 23 can be represented as a sum of multiples of 4 and 9. Hence all primes except 2,3,5,7,11,19,23 are in this sequence. [Charles R Greathouse IV, Apr 19 2010]

Examples

			A000040(11) = 31 = 3^2 + 3^2 + 3^2 + 2^2, therefore 31 is a term.
		

Crossrefs

Programs

  • Mathematica
    Join[{13,17},Prime[Range[10,100]]] (* Harvey P. Dale, May 12 2014 *)
  • PARI
    a(n)=if(n<3,[13,17][n],prime(n+7))

Extensions

Comments, reference, and links by Charles R Greathouse IV, Apr 19 2010

A298640 Number of compositions (ordered partitions) of n^2 into squares > 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 8, 12, 129, 874, 9630, 167001, 3043147, 72844510, 2423789655, 106665874384, 6156805673648, 470151743582651, 47558937432498729, 6363358599941131580, 1126147544855148769425, 263646401550138303553708, 81649922556593759124887197
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 24 2018

Keywords

Examples

			a(5) = 8 because we have [25], [16, 9], [9, 16], [9, 4, 4, 4, 4], [4, 9, 4, 4, 4], [4, 4, 9, 4, 4], [4, 4, 4, 9, 4] and [4, 4, 4, 4, 9].
		

Crossrefs

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
          add(b(n-j^2), j=2..isqrt(n)))
        end:
    a:= n-> b(n^2):
    seq(a(n), n=0..25);  # Alois P. Heinz, Feb 05 2018
  • Mathematica
    b[n_] := b[n] = If[n == 0, 1, Sum[b[n - j^2], {j, 2, Floor @ Sqrt[n]}]];
    a[n_] := b[n^2];
    Table[a[n], {n, 0, 25}] (* Jean-François Alcover, May 21 2018, after Alois P. Heinz *)

Formula

a(n) = [x^(n^2)] 1/(1 - Sum_{k>=2} x^(k^2)).
a(n) = A280542(A000290(n)).
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