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.

Showing 1-7 of 7 results.

A103278 Smallest j associated with a(n) in A103277.

Original entry on oeis.org

1, 36, 1200, 37800, 83160, 846720, 1965600, 15724800, 34927200, 279417600, 1437836400, 11502691200, 5751345600, 160626866400, 46010764800, 1927522396800, 8561475468000, 80173757664000
Offset: 1

Views

Author

David W. Wilson, Jan 27 2005

Keywords

Crossrefs

Cf. A103277, A006501 (max product of parts of partition into 3 parts).

Extensions

a(10)-a(15) from Donovan Johnson, Mar 29 2010
a(16)-a(18) from Duncan Moore, Sep 02 2017

A375580 a(n) is the number of partitions n = x + y + z of positive integers such that x*y*z is a perfect cube.

Original entry on oeis.org

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

Views

Author

Felix Huber, Aug 19 2024

Keywords

Comments

a(n) is also the number of distinct integer-sided cuboids with total edge length 4*n whose unit cubes can be grouped to a cube.
Conjecture: for n > 176, a(n) > 0. - Charles R Greathouse IV, Aug 20 2024

Examples

			a(21) = 3 because the three partitions [1, 4, 16], [3, 6, 12], [7, 7, 7] satisfy the conditions: 1 + 4 + 16 = 21 and 1*4*16 = 4^3, 3 + 6 + 12 = 21 and 3*6*12 = 6^3, 7 + 7 + 7 = 21 and 7*7*7 = 7^3.
See also linked Maple code.
		

Crossrefs

Programs

  • Maple
    # See Huber link.
  • PARI
    a(n)=sum(x=1,n\3, sum(y=x,(n-x)\2, ispower(x*y*(n-x-y),3))) \\ Charles R Greathouse IV, Aug 20 2024
    
  • PARI
    \\ See Corneth link
    
  • Python
    from sympy import integer_nthroot
    def A375580(n): return sum(1 for x in range(n//3) for y in range(x,n-x-1>>1) if integer_nthroot((n-x-y-2)*(x+1)*(y+1),3)[1]) # Chai Wah Wu, Aug 21 2024

Formula

Trivial upper bound: a(n) <= A069905(n). - Charles R Greathouse IV, Aug 23 2024

A060277 Number of m for which a+b+c = n; abc = m has at least two distinct solutions (a,b,c) with 1 <= a <= b <= c.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 1, 0, 1, 1, 3, 1, 1, 1, 1, 3, 2, 7, 3, 2, 5, 4, 3, 5, 9, 2, 5, 6, 9, 5, 9, 14, 9, 7, 5, 10, 10, 11, 18, 7, 11, 16, 14, 12, 12, 23, 19, 13, 18, 11, 20, 19, 32, 17, 21, 18, 25, 19, 21, 27, 22, 21, 31, 27, 24, 28, 42, 34, 33, 21, 28, 31, 35, 47
Offset: 1

Views

Author

Naohiro Nomoto, Mar 23 2001

Keywords

Comments

A triple (a,b,c) as described in the name cannot have c prime. - David A. Corneth, Aug 01 2018

Examples

			(14 = 6+6+2 = 8+3+3, 72 = 6*6*2 = 8*3*3); (14 = 8+5+1 = 10+2+2, 40 = 8*5*1 = 10*2*2); 14 has two "m" variables. so a(14)=2.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Count[ Tally[ Times @@@ IntegerPartitions[n, {3}]], {m_,c_} /; c>1]; Array[a, 84] (* Giovanni Resta, Jul 27 2018 *)
  • PARI
    a(n)={my(M=Map()); for(i=n\3, n, for(j=(n-i+1)\2, min(n-1-i, i), my(k=n-i-j); my(m=i*j*k); my(z); mapput(M, m, if(mapisdefined(M, m, &z), z + 1, 1)))); #select(z->z>=2, if(#M, Mat(M)[, 2], []))} \\ Andrew Howroyd, Jul 27 2018

Formula

a(n) = Sum_{k>=2} A317578(n,k). - Alois P. Heinz, Aug 01 2018

Extensions

Description revised by David W. Wilson and Don Reble, Jun 04 2002

A317578 Number T(n,k) of distinct integers that are product of the parts of exactly k partitions of n into 3 positive parts; triangle T(n,k), n>=3, k>=1, read by rows.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 12, 1, 12, 2, 19, 19, 1, 22, 1, 27, 28, 1, 31, 1, 31, 3, 38, 1, 42, 1, 46, 1, 50, 1, 50, 3, 57, 2, 51, 7, 64, 3, 71, 2, 70, 5, 77, 4, 85, 3, 86, 5, 84, 9, 104, 2, 104, 5, 108, 6, 108, 8, 1, 123, 5, 122, 9, 119, 14, 136, 9, 147, 7
Offset: 3

Views

Author

Alois P. Heinz, Jul 31 2018

Keywords

Examples

			T(13,2) = 1: only 36 is product of the parts of exactly 2 partitions of 13 into 3 positive parts: [6,6,1], [9,2,2].
T(14,2) = 2: 40 ([8,5,1], [10,2,2]) and 72 ([6,6,2], [8,3,3]).
T(39,3) = 1: 1200 ([20,15,4], [24,10,5], [25,8,6]).
T(49,3) = 2: 3024 ([24,18,7], [27,14,8], [28,12,9]) and 3600 ([20,20,9], [24,15,10], [25,12,12]).
Triangle T(n,k) begins:
   1;
   1;
   2;
   3;
   4;
   5;
   7;
   8;
  10;
  12;
  12, 1;
  12, 2;
  19;
  19, 1;
  22, 1;
		

Crossrefs

Row sums give A306403.
Column k=1 gives A306435.

Programs

  • Maple
    b:= proc(n) option remember; local m, c, i, j, h, w;
          m, c:= proc() 0 end, 0; forget(m);
          for i to iquo(n, 3) do for j from i to iquo(n-i, 2) do
            h:= i*j*(n-j-i);
            w:= m(h); w:= w+1; m(h):= w;
            c:= c+x^w-x^(w-1)
          od od; c
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n)):
    seq(T(n), n=3..100);
  • Mathematica
    b[n_] := b[n] = Module[{m, c, i, j, h, w} , m[_] = 0; c = 0; For[i = 1, i <= Quotient[n, 3], i++, For[j = i, j <= Quotient[n - i, 2], j++, h = i*j*(n-j-i); w = m[h]; w++; m[h] = w; c = c + x^w - x^(w-1)]]; c];
    T[n_] := CoefficientList[b[n], x] // Rest;
    T /@ Range[3, 100] // Flatten (* Jean-François Alcover, Jun 13 2021, after Alois P. Heinz *)

Formula

Sum_{k>=1} k * T(n,k) = A001399(n-3) = A069905(n) = A211540(n+2).
Sum_{k>=2} T(n,k) = A060277(n).
min { n >= 0 : T(n,k) > 0 } = A103277(k).

A119028 Numbers having at least 3 unique partitions into exactly 3 parts with the same product.

Original entry on oeis.org

39, 45, 49, 53, 62, 64, 65, 70, 71, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 87, 88, 89, 90, 91, 92, 93, 94, 95, 97, 98, 99, 100, 101, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128
Offset: 1

Views

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Jul 23 2006, Aug 10 2006

Keywords

Comments

That is, numbers j such that there exist positive integers a1 <= a2 <= a3, b1 <= b2 <= b3, c1 <= c2 <= c3 (unique as triples) with j = a1 + a2 + a3 = b1 + b2 + b3 = c1 + c2 + c3 and a1*a2*a3 = b1*b2*b3 = c1*c2*c3. The answer to a question raised by Tanya Khovanova, Jul 23 2006.
All integers >= 103 are members of this sequence: see second comment in A103277. - Charles Kluepfel and M. F. Hasler, Nov 23 2018

Examples

			49 = 7 + 18 + 24    7*18*24 = 3024
49 = 8 + 14 + 27    8*14*27 = 3024
49 = 9 + 12 + 28    9*12*28 = 3024
or
49 =  9 + 20 + 20   9*20*20 = 3600
49 = 10 + 15 + 24  10*15*24 = 3600
49 = 12 + 12 + 25  12*12*25 = 3600
		

Crossrefs

Programs

  • Mathematica
    pdt[lst_] := lst[[1]]*lst[[2]]*lst[[3]];
    tanya[n_] := Max[Length /@ Split[Sort[pdt /@ Union[ Partition[Last /@ Flatten[ FindInstance[a + b + c == n && a >= b >= c > 0, {a, b, c}, Integers,(* failsafe *) PartitionsP@n]], 3]] ]]];
    Select[ Range[4, 121], tanya@# >= 3 (*or strictly = ?*) &]
    Select[Range[3, 121], Max[Length /@ Split[Sort[Times @@@ Partition[Last /@ Flatten[FindInstance[a + b + c == # && a >= b >= c > 0, {a, b, c}, Integers,(* cf A069905 *) Round[ #^2/12]]], 3]]]] >= 3 &]

Extensions

More terms from Robert G. Wilson v, Jul 27 2006

A119646 a(n) = maximum number of partitions of n into 3 parts, each having the same product.

Original entry on oeis.org

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

Views

Author

Joseph Biberstine (jrbibers(AT)indiana.edu) and Robert G. Wilson v, Jul 27 2006

Keywords

Examples

			a(3)=1, because there is only one way to partition 3.
a(13)=2, because 13 = 6+6+1 = 9+2+2 and 6*6*1 = 9*2*2 = 36.
a(39)=3, because 39 = 20+15+4 = 24+10+5 = 25+8+6 and 20*15*4 = 24*10*5 = 25*8*6 = 1200.
See A103277 for more examples.
		

Crossrefs

Programs

  • Mathematica
    pdt[lst_] := lst[[1]]*lst[[2]]*lst[[3]];
    tanya[n_] := Max[Length /@ Split[Sort[pdt /@ Union[ Partition[Last /@ Flatten[ FindInstance[a + b + c == n && a >= b >= c > 0, {a, b, c}, Integers,(* failsafe *) Round[n^2/12]]], 3]] ]]];
    Table[ tanya@n, {n, 4, 108}]
    Table[SortBy[Tally[Times@@@IntegerPartitions[n,{3}]],Last][[-1,2]],{n,3,110}] (* Harvey P. Dale, Jan 08 2023 *)

Extensions

Name clarified by Dmitry Kamenetsky, Aug 02 2015

A321993 Numbers having less than 3 unique partitions into exactly 3 parts with the same product: complement of A119028.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 63, 66, 67, 68, 69, 72, 73, 80, 86, 96, 102
Offset: 1

Views

Author

M. F. Hasler, Nov 23 2018

Keywords

Comments

It is known that A119028 contains all integers >= 103, see second comment in A103277.

Crossrefs

Showing 1-7 of 7 results.