A048297 -id:A048297 - cp's OEIS frontend

A003328 Numbers that are the sum of 5 positive cubes.

5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194, 195, 199

Offset: 1

N. J. A. Sloane

As the order of addition doesn't matter we can assume terms are in increasing order. - David A. Corneth, Aug 01 2020

It seems only a finite number N of positive integers are not in this sequence, and thus a(n) = n - N for all sufficiently large n. Is it true that 2243453, last term of A048927, is sufficiently large in that sense? - M. F. Hasler, Jan 04 2023

			From _David A. Corneth_, Aug 01 2020: (Start)
3084 is in the sequence as 3084 = 5^3 + 5^3 + 5^3 +  8^3 + 13^3.
4385 is in the sequence as 4385 = 4^3 + 4^3 + 9^3 + 11^3 + 13^3.
5426 is in the sequence as 5426 = 8^3 + 9^3 + 9^3 + 12^3 + 12^3. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Cubic Number.

Cf. A003328, A048926, A048297.
Cf. A057906 (Complement)
Cf. A###### (x, y) = Numbers that are the sum of x nonzero y-th powers:
  A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).

select( {is_A003328(n,k=5,m=3,L=sqrtnint(abs(n-k+1),m))=if( n>k*L^m || nM. F. Hasler, Aug 02 2020
A003328_upto(N,k=5,m=3)=[i|i<-[1..#N=sum(n=1,sqrtnint(N,m),'x^n^m,O('x^N))^k], polcoef(N,i)] \\ M. F. Hasler, Aug 02 2020

from collections import Counter
from itertools import combinations_with_replacement as combs_w_rep
def aupto(lim):
  s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
  s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_w_rep(s, 5)))
  s2counts = Counter(s2)
  return sorted(k for k in s2counts)
print(aupto(200)) # Michael S. Branicky, May 12 2021

A134055 a(n) = Sum_{k=1..n} C(n-1,k-1) * S2(n,k) for n>0, a(0)=1, where S2(n,k) = A048993(n,k) are Stirling numbers of the 2nd kind.

Original entry on oeis.org

1, 1, 2, 8, 41, 252, 1782, 14121, 123244, 1169832, 11960978, 130742196, 1518514076, 18645970943, 241030821566, 3268214127548, 46338504902485, 685145875623056, 10538790233183702, 168282662416550040, 2784205185437851772, 47646587512911994120

Offset: 0

Paul D. Hanna, Oct 08 2007

nonn

			O.g.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 41*x^4 + 252*x^5 + 1782*x^6 + 14121*x^7 +...
where
A(x) = 1 + x/(1-x)*exp(-x/(1-x)) + 2^2*x^2/(1-2*x)^2*exp(-2*x/(1-2*x))/2! + 3^3*x^3/(1-3*x)^3*exp(-3*x/(1-3*x))/3! + 4^4*x^4/(1-4*x)^4*exp(-4*x/(1-4*x))/4! +...
simplifies to a power series in x with integer coefficients.
Illustrate the definition of the terms by:
a(4) = 1*1 + 3*7 + 3*6 + 1*1 = 41;
a(5) = 1*1 + 4*15 + 6*25 + 4*10 + 1*1 = 252;
a(6) = 1*1 + 5*31 + 10*90 + 10*65 + 5*15 + 1*1 = 1782.

Alois P. Heinz, Table of n, a(n) for n = 0..518

Cf. A037011, A048297, A048993, A122455, A218667, A218670, A245059, A245060.

a:= proc(n) option remember; local b; b:=
      proc(h, m) option remember; `if`(h=0,
        binomial(n-1, m-1), m*b(h-1, m)+b(h-1, m+1) )
      end; b(n, 0)
    end:
seq(a(n), n=0..22);  # Alois P. Heinz, Jun 24 2023

Flatten[{1,Table[Sum[Binomial[n-1,k-1] * StirlingS2[n,k],{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Aug 11 2014 *)

a(n)=if(n==0,1,sum(k=1, n, binomial(n-1, k-1)*polcoeff(1/prod(i=0, k, 1-i*x +x*O(x^(n-k))), n-k)))

a(n)=polcoeff(sum(k=0,n+1,(k*x)^k/(1-k*x)^k*exp(-k*x/(1-k*x+x*O(x^n)))/k!),n)
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Nov 04 2012

O.g.f.: Sum_{n>=0} (n*x)^n/(1-n*x)^n * exp(-n*x/(1-n*x)) / n!. - Paul D. Hanna, Nov 04 2012
From Alois P. Heinz, Jun 24 2023: (Start)
a(n) mod 2 = A037011(n) for n >= 1.
a(n) mod 2 = 1 <=> n in { A048297 } or n = 0. (End)

An initial '1' was added and definition changed slightly by Paul D. Hanna, Nov 04 2012

A088698 Replace 1 with 11 in binary representation of n.

Original entry on oeis.org

0, 3, 6, 15, 12, 27, 30, 63, 24, 51, 54, 111, 60, 123, 126, 255, 48, 99, 102, 207, 108, 219, 222, 447, 120, 243, 246, 495, 252, 507, 510, 1023, 96, 195, 198, 399, 204, 411, 414, 831, 216, 435, 438, 879, 444, 891, 894, 1791, 240, 483, 486, 975, 492, 987, 990

Offset: 0

Ralf Stephan, Oct 07 2003

			n=9: 1001 -> 110011 = 51, so a(9) = 51.

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Ordered terms plus one are in A048297.
Same sequence sorted into ascending order: A277335, A290258 (without 0).
Cf. A001196, A084471, A048678, A088697.
Main diagonal of A341520, right edge of A341521.

a(n)=if(n<1,0,if(n%2==0,2*a(n/2),4*a((n-1)/2)+3))

def a(n): return int(bin(n)[2:].replace('1', '11'), 2)
print([a(n) for n in range(55)]) # Michael S. Branicky, Feb 20 2021

a(0)=0, a(2n) = 2a(n), a(2n+1) = 4a(n) + 3.

A327616 Irregular table read by rows; the first row contains a single 1; for any n > 1, row n+1 corresponds to the ordinal transform of the terms in rows 1..n.

Original entry on oeis.org

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

Offset: 1

Rémy Sigrist, Jul 06 2020

The ordinal transform of a sequence b(n) is the sequence t(n) = number of values in b(1), ..., b(n) which are equal to b(n).

			Triangle begins:
  1:  [1]
  2:  [1]
  3:  [1, 2]
  4:  [1, 2, 3, 1]
  5:  [1, 2, 3, 1, 4, 2, 1, 5]
  6:  [1, 2, 3, 1, 4, 2, 1, 5, 6, 3, 2, 7, 1, 4, 8, 1]

Rémy Sigrist, Table of n, a(n) for n = 1..8192

Cf. A000045, A011782 (row lengths), A048297, A060138, A107946.

{ for (n=1, #a=vector(85), print1 (a[n]=if (n==1, a[n]=1, a[n]=o[a[k++]]++)", "); if (hammingweight(n)==1, k=0; o=vector(vecmax(a[1..n])))) }

Apparently:
- the greatest term in row n is A000045(n),
- a(n) = 1 iff n = 1 or n belongs to A060138,
- a(A048297(n+1)) = n (and this corresponds to the first occurrence of n),
- a(4^k) = A000045(2*k+1) for any k >= 0,
- a(2*4^k-1) = A000045(2*k+2) for any k >= 0.

A057227 Smallest member of smallest set S(n) of positive integers containing n which satisfies "k is in S, iff 2k-1 is in S, iff 4k is in S".

Original entry on oeis.org

1, 2, 2, 1, 2, 6, 1, 2, 2, 10, 6, 2, 1, 14, 2, 1, 2, 18, 10, 2, 6, 22, 2, 6, 1, 26, 14, 1, 2, 30, 1, 2, 2, 34, 18, 2, 10, 38, 2, 10, 6, 42, 22, 6, 2, 46, 6, 2, 1, 50, 26, 1, 14, 54, 1, 14, 2, 58, 30, 2, 1, 62, 2, 1, 2, 66, 34, 2, 18, 70, 2, 18, 10, 74, 38, 10, 2, 78, 10, 2, 6, 82, 42, 6, 22

Offset: 1

Henry Bottomley, Nov 23 2000

nonn

Ratio of number of times 2 appears to number of times 1 appears tends towards (1+sqrt(5))/2. Ratio of number of times 2 appears to number of times 4m+2 appears tends towards ((1+sqrt(5))/2)^log_2(2m+1).

			a(23)=2 since if 23 is in S(23) then so are 12, 3 and 2; however 1 need not be and so the smallest member of S(23) is 2.

Michael Gilleland, Some Self-Similar Integer Sequences

All values of a(n) are 1 or of form 4m+2, i.e., A016825. a(n)=1 iff n is in A048297.

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A003328 Numbers that are the sum of 5 positive cubes.

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A134055 a(n) = Sum_{k=1..n} C(n-1,k-1) * S2(n,k) for n>0, a(0)=1, where S2(n,k) = A048993(n,k) are Stirling numbers of the 2nd kind.

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A088698 Replace 1 with 11 in binary representation of n.

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A327616 Irregular table read by rows; the first row contains a single 1; for any n > 1, row n+1 corresponds to the ordinal transform of the terms in rows 1..n.

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A057227 Smallest member of smallest set S(n) of positive integers containing n which satisfies "k is in S, iff 2k-1 is in S, iff 4k is in S".

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