A011371 -id:A011371 - cp's OEIS frontend

A120738 a(n) = 4*n - A000120(n).

0, 3, 7, 10, 15, 18, 22, 25, 31, 34, 38, 41, 46, 49, 53, 56, 63, 66, 70, 73, 78, 81, 85, 88, 94, 97, 101, 104, 109, 112, 116, 119, 127, 130, 134, 137, 142, 145, 149, 152, 158, 161, 165, 168, 173, 176, 180, 183, 190, 193, 197, 200, 205, 208, 212, 215, 221, 224, 228

Offset: 0

Paul Barry, Jun 29 2006

Partial sums of A090739.
a(n) is also the increasing sequence of exponents of x in Product_{k > 1} (1 + x^(2^k - 1)). - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008
Related to partial sums of the Ruler sequence A001511 by a(n) = A005187(2n), therefore {a(n)+1} are the indices of 1's in A252488. - M. F. Hasler, Jan 22 2015

Michel Marcus, Table of n, a(n) for n = 0..10000
Keith Johnson, and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.

Cf. A000120, A000225, A001316, A001511, A005187, A011371, A030308, A061549, A067624, A086343, A090739, A117972, A252488.

A120738:= func< n | 4*n-(&+Intseq(n, 2)) >;
[A120738(n): n in [0..100]]; // G. C. Greubel, Oct 20 2024

a:=n->simplify(log[2](16^n/(add(modp(binomial(n,k),2),k=0..n))));
a:=n->simplify(log[2](16^n/(2^(n-(padic[ordp](n!,2)))))); # Note: n-(padic[ordp](n!,2)) is the number of 1's in the binary expansion of n. - Paul Pearson (ppearson(AT)rochester.edu), Aug 06 2008

Table[4 n - DigitCount[n, 2, 1], {n, 0, 58}] (* Michael De Vlieger, Nov 06 2016 *)

{a(n) = if( n < 0, 0, 4*n - subst( Pol( binary( n ) ), x, 1) ) } /* Michael Somos, Aug 28 2007 */

a(n) = 4*n - hammingweight(n); \\ Michel Marcus, Nov 06 2016

# Python 3.10
def A120738(n): return (n<<2)-n.bit_count() # Chai Wah Wu, Jul 12 2022

A120738 = lambda n: 4*n - sum(n.digits(2))
print([A120738(n) for n in (0..58)]) # Peter Luschny, Nov 06 2016

a(n) = log_2(16^n/A001316(n)). [This was the original definition.]
a(n) = 2n + A005187(n).
a(n) = 3n + A011371(n).
a(n) = 4n - log_2(A001316(n)).
a(n) = log_2(A061549(n)).
2^a(n) = 16^n/A001316(n) = A061549(n).
a(n) = A086343(n) + A001511(n) for n>0. - Alford Arnold, Mar 23 2009
2^a(n) = abs(A067624(n)/A117972(n)). - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_{k>=0} (A030308(n,k)*A000225(k+2)). - Philippe Deléham, Oct 16 2011
a(n) = A005187(2n). - M. F. Hasler, Jan 22 2015

Definition simplified by M. F. Hasler, Dec 29 2012

A233272 a(n) = n + 1 + number of nonleading zeros in binary representation of n (A080791).

Original entry on oeis.org

1, 2, 4, 4, 7, 7, 8, 8, 12, 12, 13, 13, 15, 15, 16, 16, 21, 21, 22, 22, 24, 24, 25, 25, 28, 28, 29, 29, 31, 31, 32, 32, 38, 38, 39, 39, 41, 41, 42, 42, 45, 45, 46, 46, 48, 48, 49, 49, 53, 53, 54, 54, 56, 56, 57, 57, 60, 60, 61, 61, 63, 63, 64, 64, 71, 71, 72

Offset: 0

Antti Karttunen, Dec 12 2013

From Antti Karttunen, Jan 30 2022: (Start)
Write n in binary: 1ab..xyz, then a(n) = (1+1ab..xy) + (1+1ab..x) + ... + (1+1ab) + (1+1a) + (1+1) + (1+0) + 1. This method was found by LODA miner, see the assembly program at C. Krause link.
Proof: Compare to a similar formula given for A011371, with a(n) = a(floor(n/2)) + floor(n/2) to the new formula for this sequence which is a(n) = 1 + a(floor(n/2)) + floor(n/2), for n > 0 and a(0) = 1. It is easy to see that the difference between these, a(n) - A011371(n) = 1+A070939(n), for n > 0. As A011371(n) = n minus (number of 1's in binary expansion of n), then a(n) = 1 + (number of digits in binary expansion of n) + (n minus number of 1's in binary expansion of n) = 1 + n + (number of nonleading 0's in binary expansion of n), which indeed is the definition of this sequence.
(End)

Antti Karttunen, Table of n, a(n) for n = 0..8192
Christian Krause, et al, A mined LODA assembly source for this sequence
Index entries for sequences related to binary expansion of n

Bisection: A233273 (A120511 + 1). Iterates: A233271, A216431.

Cf. A000120, A005187, A011371, A054429, A070939, A080791, A092391.

DigitCount[#, 2, 0] + # + 1 & [Range[0, 100]] (* Paolo Xausa, Mar 01 2024 *)

A233272(n) = { my(s=1); while(n, n>>=1; s+=(1+n)); (s); }; \\ (After a LODA-assembly program found by a miner) - Antti Karttunen, Jan 30 2022

(define (A233272 n) (+ 1 n (A080791 n)))
;; Alternatively:
(define (A233272 n) (if (zero? n) 1 (+ n (A000120 (A054429 n)))))

a(n) = n + A080791(n) + 1.
For all n>=1, a(n) = n + A000120(A054429(n)).
a(0) = 1; for n > 1, a(n) = 1 + floor(n/2) + a(floor(n/2)). - (Found by LODA miner, see comments) - Antti Karttunen, Jan 30 2022

A325509 Number of factorizations of n! into factorial numbers > 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Gus Wiseman, May 08 2019

nonn

			n = 10:
  (6*120*5040)
  (720*5040)
  (3628800)
n = 16:
  (2*2*2*2*1307674368000)
  (2*120*87178291200)
  (20922789888000)
n = 24:
  (2*2*6*25852016738884976640000)
  (24*25852016738884976640000)
  (620448401733239439360000)

Cf. A000142, A011371, A022559, A034876, A034878, A076934, A109100, A109101, A109102, A115627, A135291, A322583, A325272, A325273, A325276, A325508, A325510, A325511.

facs[n_,u_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d,u],Min@@#>=d&]],{d,Intersection[u,Rest[Divisors[n]]]}]];
Table[Length[facs[n!,Rest[Array[#!&,n]]]],{n,15}]

a(n) = 1 + A034876(n).

More terms from Alois P. Heinz, May 08 2019

A257265 Distance to n from a leaf nearest to n in the binary beanstalk.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Apr 29 2015

nonn

If n is one of the terms of A055938 (that is, one of the leaves of the binary beanstalk), then a(n) is zero. Otherwise, a(n) is the least number of iterations of A011371 required to reach n, when we start from any term of A055938.
Compare to the definition of A213725, which gives 1 + distance to the farthest leaf in the binary beanstalk (giving 0 for the nodes which reside in A179016, the infinite stem of the beanstalk, as there the maximum distance would not have a finite value).
Note that although the recursive formula given here mirrors the one given at A213725, it cannot be implemented in a naive way, precisely because of that infinite stem, as it would lead to recursion without end. Instead, with ordinary eagerly evaluating programming languages we have to employ, for example, a breadth-first search, as in the given Scheme-program.
Question: Will there ever appear a term larger than 2? (Only terms 0 - 2 occur in range 0 .. 2097151).

			For 0, the nearest leaf is 2, as when we start from 2, and always subtract the binary weight, A000120, we have: 2 - A000120(2) = A011371(2) = 1, and A011371(1) = 0, thus it takes two steps to get to 0, and there are no other terms of A055938 from which it would take fewer steps), so a(0) = 2, and also a(1) = 1, because it's one step nearer to 2.
a(2) = 0, because 2 is one of the terms of A055938.
a(8) = 2, because 12, 13 and 14 are the three nearest leaves to 8, and A011371(12) = A011371(13) = 10, A011371(14) = 11, A011371(10) = A011371(11) = 8 (thus it takes two iterations of A011371 to reach 8 from any of those three leaves) and there are no leaves nearer.
Please see also Paul Tek's illustration.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Ela, peano
Haskell Wiki, Peano numbers
Paul Tek, Illustration of how natural numbers in range 0 .. 133 are organized as a binary tree in the binary beanstalk

Cf. A055938 (positions of 0's), A257508 (of 1's), A257509 (of 2's).
Cf. also A000120, A011371, A079559, A213723, A213724.
Cf. also A179016, A213725, A257264.

a257265 = length . us where
   us n = if a079559 n == 0
             then [] else () : zipWith (const $ const ())
                               (us $ a213723 n) (us $ a213724 n)
-- Reinhard Zumkeller, May 03 2015

;; Do a breadth-first search over the descendants, which are at each step of iteration sorted by their distance from the starting node.
(define (A257265 n) (let loop ((descendants (list (cons 0 n)))) (let ((dist (caar descendants)) (node (cdar descendants))) (cond ((zero? (A079559 node)) dist) (else (loop (sort (append (list (cons (+ 1 dist) (A213724 node)) (cons (+ 1 dist) (A213723 node))) (cdr descendants)) (lambda (a b) (< (car a) (car b))))))))))

If A079559(n) = 0, then a(n) = 0, otherwise a(n) = 1 + min(a(A213723(n)), a(A213724(n))). [But please see the comments above.]

A336496 Products of superfactorials (A000178).

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 24, 32, 48, 64, 96, 128, 144, 192, 256, 288, 384, 512, 576, 768, 1024, 1152, 1536, 1728, 2048, 2304, 3072, 3456, 4096, 4608, 6144, 6912, 8192, 9216, 12288, 13824, 16384, 18432, 20736, 24576, 27648, 32768, 34560, 36864, 41472, 49152, 55296

Offset: 1

Gus Wiseman, Aug 03 2020

nonn

First differs from A317804 in having 34560, which is the first term with more than two distinct prime factors.

			The sequence of terms together with their prime indices begins:
    1: {}
    2: {1}
    4: {1,1}
    8: {1,1,1}
   12: {1,1,2}
   16: {1,1,1,1}
   24: {1,1,1,2}
   32: {1,1,1,1,1}
   48: {1,1,1,1,2}
   64: {1,1,1,1,1,1}
   96: {1,1,1,1,1,2}
  128: {1,1,1,1,1,1,1}
  144: {1,1,1,1,2,2}
  192: {1,1,1,1,1,1,2}
  256: {1,1,1,1,1,1,1,1}
  288: {1,1,1,1,1,2,2}
  384: {1,1,1,1,1,1,1,2}
  512: {1,1,1,1,1,1,1,1,1}

A001013 is the version for factorials, with complement A093373.
A181818 is the version for superprimorials, with complement A336426.
A336497 is the complement.
A000178 lists superfactorials.
A001055 counts factorizations.
A006939 lists superprimorials or Chernoff numbers.
A049711 is the minimum prime multiplicity in A000178.
A174605 is the maximum prime multiplicity in A000178.
A303279 counts prime factors of superfactorials.
A317829 counts factorizations of superprimorials.
A322583 counts factorizations into factorials.
A325509 counts factorizations of factorials into factorials.
Cf. A000142, A000720, A007489, A011371, A022559, A022915, A027423, A034878, A034876, A076954, A115627, A294068.

supfac[n_]:=Product[k!,{k,n}];
facsusing[s_,n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facsusing[Select[s,Divisible[n/d,#]&],n/d],Min@@#>=d&]],{d,Select[s,Divisible[n,#]&]}]];
Select[Range[1000],facsusing[Rest[Array[supfac,30]],#]!={}&]

A001801 Coefficients of Legendre polynomials.

Original entry on oeis.org

3, 15, 105, 315, 6930, 18018, 90090, 218790, 2078505, 4849845, 22309287, 50702925, 1825305300, 4071834900, 18032411700, 39671305740, 347123925225, 755505013725, 3273855059475, 7064634602025, 121511715154830, 260382246760350, 1112542327066950, 2370198870707850, 20146690401016725

Offset: 0

N. J. A. Sloane

nonn

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Milan Janjic, Some classes of numbers and derivatives, JIS 12 (2009) 09.8.3.

Cf. A001790, A001796, A008316, A011371.
Bisection of A004733.
Diagonal 3 of triangle A100258.

A001801:= func< n | 3*Binomial(n+3,3)*Catalan(n+2)*2^(Valuation(Factorial(n+4),2)-n-4) >;
[A001801(n): n in [0..30]]; // G. C. Greubel, Apr 26 2025

A001801[n_]:= 3*2^(2*n+1)*Binomial[n+3/2, n]/2^DigitCount[n+4,2,1];
Table[A001801[n], {n,0,40}] (* G. C. Greubel, Apr 26 2025 *)

a(n)=if(n<0,0,polcoeff(pollegendre(n+4),n)*2^valuation((n\2*2+4)!,2))

def A001801(n): return 3*2^(n-3)*binomial(n+3/2,n)*2^valuation(factorial(n+4), 2)
print([A001801(n) for n in range(31)]) # G. C. Greubel, Apr 26 2025

a(n) = 3*2^(n-3)*binomial(n + 3/2, n)*2^A011371(n+4). - G. C. Greubel, Apr 26 2025

More terms from Michael Somos, Oct 25 2002

A013937 a(n) = Sum_{k=1..n} floor(n/k^3).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 75, 76, 77, 78, 79, 80, 81, 82

Offset: 0

N. J. A. Sloane, Henri Lifchitz

nonn

			a(36) = [36/1]+[36/8]+[36/27]+[36/64]+... = 36+4+1+0+... = 41.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Benoit Cloitre, Plot of (a(n)-zeta(3)*n)/n^(1/3)-zeta(1/3)

Cf. A005187, A006218, A011371, A013936, A013939 for similar sequences.

A013937:=n->add(floor(n/k^3), k=1..n); seq(A013937(n), n=0..100); # Wesley Ivan Hurt, Feb 15 2014

Table[Sum[Floor[n/k^3], {k, n}], {n, 0, 100}] (* Wesley Ivan Hurt, Feb 15 2014 *)

a(n)=sum(k=1,ceil(n^(1/3)),n\k^3) /*Benoit Cloitre, Nov 05 2012 */

a(n) = a(n-1)+A061704(n). a(n) = Sum_{k=1..n} floor((n/k)^(1/3)) with asymptotic formula: a(n) = zeta(3)*n+zeta(1/3)*n^(1/3)+O(n^theta) where theta<1/3 and we conjecture that theta=1/4+epsilon is the best possible choice. - Benoit Cloitre, Nov 05 2012

G.f.: (1/(1 - x))*Sum_{k>=1} x^(k^3)/(1 - x^(k^3)). - Ilya Gutkovskiy, Feb 11 2017

More terms from Henry Bottomley, Jul 03 2001

A054893 a(n) = Sum_{j > 0} floor(n/4^j).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, May 23 2000

Different from highest power of 4 dividing n! (see A090616).

			  a(10^0) = 0.
  a(10^1) = 2.
  a(10^2) = 32.
  a(10^3) = 330.
  a(10^4) = 3331.
  a(10^5) = 33330.
  a(10^6) = 333330.
  a(10^7) = 3333329.
  a(10^8) = 33333328.
  a(10^9) = 333333326.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A053737, A235127 (first differences).

Cf. A011371, A027868, A054861, A054899, A067080, A090616, A098844, A132028.

function A054893(n)
  if n eq 0 then return n;
  else return A054893(Floor(n/4)) + Floor(n/4);
  end if; return A054893;
end function;
[A054893(n): n in [0..103]]; // G. C. Greubel, Feb 09 2023

Table[t=0; p=4; While[s=Floor[n/p]; t=t+s; s>0, p *= 4]; t, {n,0,100}]
Table[Total[Floor/@(n/NestList[4#&,4,6])],{n,0,80}] (* Harvey P. Dale, Jun 12 2022 *)

a(n) = (n - sumdigits(n,4))/3; \\ Kevin Ryde, Jan 08 2024

def A054893(n):
    if (n==0): return 0
    else: return A054893(n//4) + (n//4)
[A054893(n) for n in range(104)] # G. C. Greubel, Feb 09 2023

a(n) = floor(n/4) + floor(n/16) + floor(n/64) + floor(n/256) + ...
a(n) = (n - A053737(n))/3.
From Hieronymus Fischer, Sep 15 2007: (Start)
a(n) = a(floor(n/4)) + floor(n/4).
a(4*n) = a(n) + n.
a(n*4^m) = a(n) + n*(4^m-1)/3.
a(k*4^m) = k*(4^m-1)/3, for 0 <= k < 4, m >= 0.
Asymptotic behavior:
a(n) = n/3 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/3; equality holds true for powers of 4.
a(n) >= (n-3)/3 - floor(log_4(n)); equality holds true for n = 4^m - 1, m>0. lim inf (n/3 - a(n)) = 1/3, for n-->oo.
lim sup (n/3 - log_4(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_4(n)) = 0, for n-->oo.
G.f.: (1/(1-x))*Sum_{k > 0} x^(4^k)/(1-x^(4^k)). (End)
Partial sums of A235127. - R. J. Mathar, Jul 08 2021

Edited by Hieronymus Fischer, Sep 15 2007

Examples added by Hieronymus Fischer, Jun 06 2012

A065040 Triangle read by rows: T(m,k) = exponent of the highest power of 2 dividing the binomial coefficient binomial(m,k).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 2, 3, 1, 3, 2, 3, 0, 0, 0, 2, 2, 1, 1, 2, 2, 0, 0, 0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0

Offset: 0

Claude Lenormand (hlne.lenormand(AT)voono.net), Nov 05 2001

T(m,k) is the number of 'carries' that occur when adding k and m-k in base 2 using the traditional addition algorithm. - Tom Edgar, Jun 10 2014

			Triangle begins:
[0]
[0, 0]
[0, 1, 0]
[0, 0, 0, 0]
[0, 2, 1, 2, 0]
[0, 0, 1, 1, 0, 0]
[0, 1, 0, 2, 0, 1, 0]
[0, 0, 0, 0, 0, 0, 0, 0]
[0, 3, 2, 3, 1, 3, 2, 3, 0]
[0, 0, 2, 2, 1, 1, 2, 2, 0, 0]
[0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0]
... - _N. J. A. Sloane_, Aug 21 2021

Antti Karttunen, Table of n, a(n) for n = 0..10010; the first 141 antidiagonals, flattened
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Row sums: A187059.

Cf. A007318, A007814, A001511, A000120, A047999, A049606, A000680, A048881, A011371, A005187, A000265, A001316, A001317, A243759.

A065040 := (n, k) -> padic[ordp](binomial(n, k), 2):
seq(seq(A065040(n,k), k=0..n), n=0..13); # Peter Luschny, Aug 15 2017

T[m_, k_] := IntegerExponent[Binomial[m, k], 2]; Table[T[m, k], {m, 0, 13}, {k, 0, m}] // Flatten (* Jean-François Alcover, Oct 06 2016 *)

T(m,k)=hammingweight(k)+hammingweight(m-k)-hammingweight(m)
for(m=0,9,for(k=0,m,print1(T(m,k)", "))) \\ Charles R Greathouse IV, Mar 26 2013

As an array f(i,j) = f(j,i) = T(i+j,j) read by antidiagonals: f(0,j) = 0, f(1,j) = A007814(j+1), f(i,j) = Sum_{k=0..i-1} (f(1,j+k) - f(1,k)). [corrected by Kevin Ryde, Oct 07 2021]
The n-th term a(n) is equal to the binomial coefficient binomial(m,k), where m = floor((1+sqrt(8*n+1))/2) - 1 and k = n - m(m+1)/2. Also a(n) = g(m) - g(k) - g(m-k), where g(x) = Sum_{i=1..floor(log_2(x))} floor(x/2^i), m = floor((1+sqrt(8*n+1))/2) - 1, k = n - m(m+1)/2. - Hieronymus Fischer, May 05 2007
T(m,k) <= log_2 m, for m > 0. - Charles R Greathouse IV, Mar 26 2013
T(m,k) = log_2(A082907(m,k)). - Tom Edgar, Jun 10 2014
From Antti Karttunen, Oct 28 2014: (Start)
a(n) = A007814(A007318(n)).
a(n) * A047999(n) = 0 and a(n) + A047999(n) > 0 for all n.
(End)

Name clarified by Antti Karttunen, Oct 28 2014

A085604 T(n,k) = highest power of prime(k) dividing n!, read by rows.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 3, 1, 0, 0, 3, 1, 1, 0, 0, 4, 2, 1, 0, 0, 0, 4, 2, 1, 1, 0, 0, 0, 7, 2, 1, 1, 0, 0, 0, 0, 7, 4, 1, 1, 0, 0, 0, 0, 0, 8, 4, 2, 1, 0, 0, 0, 0, 0, 0, 8, 4, 2, 1, 1, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 10, 5, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 11, 5, 2, 2, 1, 1, 0, 0, 0

Offset: 1

Reinhard Zumkeller, Jul 07 2003

T(n,1) = A011371(n); T(n,2) = A054861(n) for n>1;
T(n,k) = number of occurrences of prime(k) as factor in numbers <= n (with repetitions);
Sum{T(n,k): 1<=k<=n} = A022559(n);
T(n, A000720(n)) = 1; T(n,k) = 0, A000720(n)T(n,k) = A115627(n,k) for n > 1 and k=1..A000720(n). - Reinhard Zumkeller, Nov 01 2013

			0;
1,0;
1,1,0;
3,1,0,0;
3,1,1,0,0;
4,2,1,0,0,0;
4,2,1,1,0,0,0;
7,2,1,1,0,0,0,0;
7,4,1,1,0,0,0,0,0;
8,4,2,1,0,0,0,0,0,0;

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A141809, A115627, A000142.

a085604 n k = a085604_tabl !! (n-2) !! (k-1)
a085604_row 1 = [0]
a085604_row n = a115627_row n ++ (take $ a062298 $ fromIntegral n) [0,0..]
a085604_tabl = map a085604_row [1..]
-- Reinhard Zumkeller, Nov 01 2013

T[n_, k_] := Module[{p = Prime[k], jm}, jm = Floor[Log[p, n]]; Sum[Quotient[n, p^j], {j, 1, jm}]];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 19 2021 *)

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