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 1-10 of 16 results. Next

A316413 Heinz numbers of integer partitions whose length divides their sum.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 16, 17, 19, 21, 22, 23, 25, 27, 28, 29, 30, 31, 32, 34, 37, 39, 41, 43, 46, 47, 49, 53, 55, 57, 59, 61, 62, 64, 67, 68, 71, 73, 78, 79, 81, 82, 83, 84, 85, 87, 88, 89, 90, 91, 94, 97, 98, 99, 100, 101, 103, 105, 107, 109, 110
Offset: 1

Views

Author

Gus Wiseman, Jul 02 2018

Keywords

Comments

In other words, partitions whose average is an integer.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			Sequence of partitions whose length divides their sum begins (1), (2), (11), (3), (4), (111), (22), (31), (5), (6), (1111), (7), (8), (42), (51), (9), (33), (222), (411).
		

Crossrefs

Programs

  • Maple
    isA326413 := proc(n)
        psigsu := A056239(n) ;
        psigle := numtheory[bigomega](n) ;
        if modp(psigsu,psigle) = 0 then
            true;
        else
            false;
        end if;
    end proc:
    n := 1:
    for i from 2 to 3000 do
        if isA326413(i) then
            printf("%d %d\n",n,i);
            n := n+1 ;
        end if;
    end do: # R. J. Mathar, Aug 09 2019
    # second Maple program:
    q:= n-> (l-> nops(l)>0 and irem(add(i, i=l), nops(l))=0)(map
            (i-> numtheory[pi](i[1])$i[2], ifactors(n)[2])):
    select(q, [$1..110])[];  # Alois P. Heinz, Nov 19 2021
  • Mathematica
    Select[Range[2,100],Divisible[Total[Cases[FactorInteger[#],{p_,k_}:>k*PrimePi[p]]],PrimeOmega[#]]&]

A074761 Number of partitions of n of order n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 9, 1, 4, 5, 1, 1, 12, 1, 27, 7, 6, 1, 81, 1, 7, 1, 54, 1, 407, 1, 1, 11, 9, 13, 494, 1, 10, 13, 423, 1, 981, 1, 137, 115, 12, 1, 1309, 1, 59, 17, 193, 1, 240, 21, 1207, 19, 15, 1, 47274, 1, 16, 239, 1, 25, 3284, 1, 333, 23, 3731, 1, 42109, 1, 19
Offset: 1

Views

Author

Vladeta Jovovic, Sep 28 2002

Keywords

Comments

Order of partition is lcm of its parts.
a(n) is the number of conjugacy classes of the symmetric group S_n such that a representative of the class has order n. Here order means the order of an element of a group. Note that a(n) = 1 if and only if n is a prime power. - W. Edwin Clark, Aug 05 2014

Examples

			The a(15) = 5 partitions are (15), (5,3,3,3,1), (5,5,3,1,1), (5,3,3,1,1,1,1), (5,3,1,1,1,1,1,1,1). - _Gus Wiseman_, Aug 01 2018
		

Crossrefs

Programs

  • Maple
    A:= proc(n)
          uses numtheory;
          local S;
        S:= add(mobius(n/i)*1/mul(1-x^j,j=divisors(i)),i=divisors(n));
        coeff(series(S,x,n+1),x,n);
    end proc:
    seq(A(n),n=1..100); # Robert Israel, Aug 06 2014
  • Mathematica
    a[n_] := With[{s = Sum[MoebiusMu[n/i]*1/Product[1-x^j, {j, Divisors[i]}], {i, Divisors[n]}]}, SeriesCoefficient[s, {x, 0, n}]]; Array[a, 80] (* Jean-François Alcover, Feb 29 2016 *)
    Table[Length[Select[IntegerPartitions[n],LCM@@#==n&]],{n,50}] (* Gus Wiseman, Aug 01 2018 *)
  • PARI
    pr(k, x)={my(t=1); fordiv(k, d, t *= (1-x^d) ); return(t); }
    a(n) =
    {
        my( x = 'x+O('x^(n+1)) );
        polcoeff( Pol( sumdiv(n, i, moebius(n/i) / pr(i, x) ) ), n );
    }
    vector(66, n, a(n) )
    \\ Joerg Arndt, Aug 06 2014

Formula

Coefficient of x^n in expansion of Sum_{i divides n} A008683(n/i)*1/Product_{j divides i} (1-x^j).

A074971 Number of partitions of n into distinct parts of order n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, 6, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 4, 1, 2, 1, 1, 1, 32, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 24, 1, 1, 1, 2, 1, 24, 1, 1, 1, 1, 1, 12, 1, 1, 1, 3, 1, 2
Offset: 1

Views

Author

Vladeta Jovovic, Oct 05 2002

Keywords

Comments

Order of partition is lcm of its parts.

Examples

			The a(36) = 6 partitions are (36), (18,12,6), (18,12,4,2), (18,12,3,2,1), (18,9,4,3,2), (12,9,6,4,3,2). - _Gus Wiseman_, Aug 01 2018
		

Crossrefs

Programs

  • PARI
    A074971(n) = { my(q=0); fordiv(n,i,my(p=1); fordiv(i,j,p *= (1 + 'x^j)); q += moebius(n/i)*p); polcoeff(q,n); }; \\ Antti Karttunen, Dec 19 2018

Formula

Coefficient of x^n in expansion of Sum_{i divides n} mu(n/i)*Product_{j divides i} (1+x^j).

A316429 Heinz numbers of integer partitions whose length is equal to their LCM.

Original entry on oeis.org

2, 6, 9, 20, 50, 56, 84, 125, 126, 176, 189, 196, 240, 294, 360, 416, 441, 540, 600, 624, 686, 810, 900, 936, 968, 1029, 1040, 1088, 1215, 1350, 1404, 1500, 1560, 2025, 2106, 2250, 2340, 2401, 2432, 2600, 2704, 3159, 3375, 3510, 3648, 3750, 3900, 4056, 5265
Offset: 1

Views

Author

Gus Wiseman, Jul 02 2018

Keywords

Comments

A110295 is a subsequence.

Examples

			3750 is the Heinz number of (3,3,3,3,2,1), whose length and lcm are both 6.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2,200],PrimeOmega[#]==LCM@@Cases[FactorInteger[#],{p_,k_}:>PrimePi[p]]&]
  • PARI
    heinz(n) = my(f=factor(n), pr=f[,1]~,exps=f[,2], res=vector(vecsum(exps)), t=0); for(i = 1, #pr, pr[i] = primepi(pr[i]); for(j=1, exps[i],t++; res[t] = pr[i])); res
    is(n) = my(h = heinz(n)); lcm(h)==#h \\ David A. Corneth, Jul 05 2018

A316430 Heinz numbers of integer partitions whose length is equal to the GCD of all the parts.

Original entry on oeis.org

1, 2, 9, 21, 39, 57, 87, 91, 111, 125, 129, 159, 183, 203, 213, 237, 247, 267, 301, 303, 321, 325, 339, 377, 393, 417, 427, 453, 489, 519, 543, 551, 553, 559, 575, 579, 597, 669, 687, 689, 707, 717, 753, 789, 791, 813, 817, 843, 845, 879, 923, 925, 933, 951, 973
Offset: 1

Views

Author

Gus Wiseman, Jul 02 2018

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
2 is the only even term in the sequence. 3k is in the sequence if and only if k is in A031215. 5k is in the sequence if and only if k = pq with p and q in A031336.

Examples

			Sequence of integer partitions whose length is equal to their GCD begins: (), (1), (2,2), (4,2), (6,2), (8,2), (10,2), (6,4), (12,2), (3,3,3), (14,2), (16,2), (18,2), (10,4), (20,2), (22,2), (8,6), (24,2), (14,4), (26,2), (28,2), (6,3,3).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[200],PrimeOmega[#]==GCD@@Cases[FactorInteger[#],{p_,k_}:>PrimePi[p]]&]
  • PARI
    is(n,f=factor(n))=gcd(apply(primepi,f[,1]))==vecsum(f[,2]) \\ Charles R Greathouse IV, Jul 25 2024

Formula

a(n) << n log^2 n, can this be improved? - Charles R Greathouse IV, Jul 25 2024

A316432 Number of integer partitions of n whose length is equal to the GCD of all parts.

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 0, 1, 1, 2, 0, 2, 0, 3, 2, 3, 0, 5, 0, 3, 4, 5, 0, 8, 1, 6, 6, 6, 0, 11, 0, 8, 10, 8, 2, 18, 0, 9, 14, 15, 0, 19, 0, 16, 21, 11, 0, 34, 1, 16, 24, 24, 0, 30, 10, 27, 30, 14, 0, 71, 0, 15, 34, 38, 18, 47, 0, 47, 44, 36, 0, 88, 0, 18, 79, 63, 5
Offset: 1

Views

Author

Gus Wiseman, Jul 02 2018

Keywords

Examples

			The a(24) = 8 partitions:
(14,10), (22,2),
(9,9,6), (12,9,3), (15,6,3), (18,3,3),
(8,8,4,4), (12,4,4,4).
		

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],GCD@@#==Length[#]&]],{n,30}]
  • PARI
    a(n) = {my(nb = 0); forpart(p=n, if (gcd(p)==#p, nb++);); nb;} \\ Michel Marcus, Jul 03 2018

A290104 a(n) = A003963(n) / A290103(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 3, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 1, 4, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 8, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 4, 2, 3, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 3, 1, 4, 1, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 13 2017

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k). Then a(n) is the product divided by the LCM of the integer partition with Heinz number n. - Gus Wiseman, Aug 01 2018

Examples

			n = 21 = 3 * 7 = prime(2) * prime(4), thus A003963(21) = 2*4 = 8, while A290103(21) = lcm(2,4) = 4, so a(21) = 8/4 = 2.
		

Crossrefs

Differs from A290106 for the first time at n=21.

Programs

  • Mathematica
    Table[If[n == 1, 1, Apply[Times, Map[PrimePi[#1]^#2 & @@ # &, #]] / Apply[LCM, PrimePi[#[[All, 1]] ]]] &@ FactorInteger@ n, {n, 120}] (* Michael De Vlieger, Aug 14 2017 *)
  • Scheme
    (define (A290104 n) (/ (A003963 n) (A290103 n)))

Formula

a(n) = A003963(n) / A290103(n).
Other identities. For all n >= 1:
a(A181819(n)) = A005361(n)/A072411(n).

A316433 Number of integer partitions of n whose length is equal to the LCM of all parts.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 2, 1, 4, 3, 4, 4, 8, 5, 7, 8, 10, 8, 13, 13, 20, 18, 25, 25, 36, 34, 48, 52, 64, 64, 85, 85, 108, 106, 129, 133, 160, 158, 189, 194, 229, 228, 276, 279, 332, 336, 394, 402, 476, 489, 572, 599, 699, 728, 845, 889, 1032, 1094, 1251, 1332, 1523
Offset: 1

Views

Author

Gus Wiseman, Jul 02 2018

Keywords

Examples

			The a(13) = 8 partitions are (4441), (55111), (322222), (332221), (333211), (622111), (631111), (7111111).
		

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],LCM@@#==Length[#]&]],{n,30}]
  • PARI
    a(n) = {my(nb = 0); forpart(p=n, if (lcm(Vec(p))==#p, nb++);); nb;} \\ Michel Marcus, Jul 03 2018

A319333 Heinz numbers of integer partitions whose sum is equal to their LCM.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 30, 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, 198, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263
Offset: 1

Views

Author

Gus Wiseman, Sep 17 2018

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			The sequence of partitions whose Heinz numbers are in the sequence begins: (1), (2), (3), (4), (5), (6), (7), (8), (9), (10), (3,2,1), (11), (12).
		

Crossrefs

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Select[Range[2,100],LCM@@primeMS[#]==Total[primeMS[#]]&]

A317624 Number of integer partitions of n where all parts are > 1 and whose LCM is n.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 5, 1, 1, 1, 17, 1, 1, 1, 7, 1, 60, 1, 1, 1, 1, 1, 76, 1, 1, 1, 55, 1, 105, 1, 11, 10, 1, 1, 187, 1, 6, 1, 13, 1, 30, 1, 111, 1, 1, 1, 5043, 1, 1, 15, 1, 1, 230, 1, 17, 1, 242, 1, 4173, 1, 1, 12, 19, 1
Offset: 0

Views

Author

Gus Wiseman, Aug 01 2018

Keywords

Examples

			The a(20) = 5 partitions are (20), (10,4,4,2), (10,4,2,2,2), (5,5,4,4,2), (5,5,4,2,2,2).
The a(45) = 10 partitions:
  (45),
  (15,15,9,3,3), (15,9,9,9,3),
  (15,9,9,3,3,3,3), (15,9,5,5,5,3,3), (9,9,9,5,5,5,3),
  (15,9,3,3,3,3,3,3,3), (9,9,5,5,5,3,3,3,3), (9,5,5,5,5,5,5,3,3),
  (9,5,5,5,3,3,3,3,3,3,3).
From _David A. Corneth_, Sep 08 2018: (Start)
Let sum(t) denote the sum of elements of a tuple t. The tuples t with distinct divisors of 45 that have lcm(t) = 45 and sum(t) <= 45 are {(45) and (3, 9, 15), (3, 5, 9, 15), (3, 5, 9), (5, 9), (9, 15), (5, 9, 15)}. For each such tuple t, find the number of partitions of 45 - s(t) into distinct parts of t.
For the tuple (45), there is 1 partition of 45 - 45 = 0 into parts with 45. That is: {()}.
For the tuple (3, 9, 15), there are 4 partitions of 45 - (3 + 9 + 15) = 18 into parts with 3, 9 and 15. They are {(3, 15), (9, 9), (3, 3, 3, 9), (3, 3, 3, 3, 3, 3)}.
For the tuple (3, 5, 9), there are 4 partitions of 45 - (3 + 5 + 9) = 28 into parts with 3, 5 and 9; they are {(5, 5, 9, 9), (3, 3, 3, 5, 5, 9), (3, 5, 5, 5, 5, 5), (3, 3, 3, 3, 3, 3, 5, 5)}.
For the tuple (3, 5, 9, 15), there is 1 partition of 45 - (3 + 5 + 9 + 15) = 13 into parts with 3, 5, 9 and 15. That is (3, 5, 5).
The other tuples, (5, 9), (9, 15), and (5, 9, 15); they give no extra tuples. That's because there is no solution to the Diophantine equation for 5x + 9y = 45 - (5 + 9), corresponding to the tuple (5, 9) with nonnegative x, y.
That also excludes (9, 15); if there is a solution for that, there would also be a solution for (5, 9). This could whittle down the number of seeds even further. Similarly, (5, 9, 15) gives no solution.
Therefore a(45) = 1 + 4 + 4 + 1 = 10.
(End)
In general, there are A318670(n) (<= A069626(n)) such seed sets of divisors where to start extending the partition from. (See the second PARI program which uses subroutine toplevel_starting_sets.) - _Antti Karttunen_, Sep 08 2018
		

Crossrefs

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],And[Min@@#>=2,LCM@@#==n]&]],{n,30}]
  • PARI
    strong_divisors_reversed(n) = vecsort(select(x -> (x>1), divisors(n)), , 4);
    partitions_into_lcm(orgn,n,parts,from=1,m=1) = if(!n,(m==orgn),my(k = #parts, s=0); for(i=from,k,if(parts[i]<=n, s += partitions_into_lcm(orgn,n-parts[i],parts,i,lcm(m,parts[i])))); (s));
    A317624(n) = if(n<=1,0,partitions_into_lcm(n,n,strong_divisors_reversed(n))); \\ Antti Karttunen, Sep 07 2018
    
  • PARI
    strong_divisors_reversed(n) = vecsort(select(x -> (x>1), divisors(n)), , 4);
    partitions_into(n,parts,from=1) = if(!n,1, if(#parts==from, (0==(n%parts[from])), my(s=0); for(i=from,#parts,if(parts[i]<=n, s += partitions_into(n-parts[i],parts,i))); (s)));
    toplevel_starting_sets(orgn,n,parts,from=1,ss=List([])) = { my(k = #parts, s=0, newss); if(lcm(Vec(ss))==orgn,s += partitions_into(n,ss)); for(i=from,k,if(parts[i]<=n, newss = List(ss); listput(newss,parts[i]); s += toplevel_starting_sets(orgn,n-parts[i],parts,i+1,newss))); (s) };
    A317624(n) = if(n<=1,0,toplevel_starting_sets(n,n,strong_divisors_reversed(n))); \\ Antti Karttunen, Sep 08-10 2018
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