A115964 -id:A115964 - cp's OEIS frontend

A024451 a(n) is the numerator of Sum_{i = 1..n} 1/prime(i).

0, 1, 5, 31, 247, 2927, 40361, 716167, 14117683, 334406399, 9920878441, 314016924901, 11819186711467, 492007393304957, 21460568175640361, 1021729465586766997, 54766551458687142251, 3263815694539731437539, 201015517717077830328949, 13585328068403621603022853

Offset: 0

N. J. A. Sloane, Clark Kimberling

Arithmetic derivative of p#: a(n) = A003415(A002110(n)). - Reinhard Zumkeller, Feb 25 2002
(n-1)-st elementary symmetric functions of first n primes; see Mathematica section. - Clark Kimberling, Dec 29 2011
Denominators of the harmonic mean of the first n primes; A250130 gives the numerators. - Colin Barker, Nov 14 2014
Let Pn(n) = A002110 denote the primorial function. The average number of distinct prime factors <= prime(n) in the natural numbers up to Pn(n) is equal to Sum_{i = 1..n} 1/prime(i). - Jamie Morken, Sep 17 2018
Conjecture: All terms are squarefree numbers. - Nicolas Bělohoubek, Apr 13 2022
The above conjecture would imply that for n > 0, gcd(a(n), A369651(n)) = 1. See corollary 2 on the page 4 of Ufnarovski-Åhlander paper. - Antti Karttunen, Jan 31 2024
Apart from the initial 0, a subsequence of A048103. Proof: For all primes p, when i >= A000720(p), neither p itself nor p^p divides a(i) [implied by Henry Bottomley's Sep 27 2006 formula], but neither does p^p divide a(i) when 0 < i < A000720(p), as then p^p > a(i). See A074107, which gives an upper bound for this sequence. - Antti Karttunen, Nov 19 2024

			0/1, 1/2, 5/6, 31/30, 247/210, 2927/2310, 40361/30030, 716167/510510, 14117683/9699690, ...

S. R. Finch, Mathematical Constants, Cambridge, 2003, Sect. 2.2.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Sect. VII.28.

Alois P. Heinz, Table of n, a(n) for n = 0..350 (terms n = 1..100 from T. D. Noe)
Victor Ufnarovski and Bo Åhlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.
Index entries for sequences related to primorial numbers

Denominators are A002110.
See also A106830/A034386, A241189/A241190, A241191/A241192, A061015/A061742, A115963/A115964, A250133/A296358, and A096795/A051451, A354417/A354418, A354859/A354860.
Row sums of A077011 and A258566.
Subsequence of A048103 (after the initial 0).
Cf. A003415, A002110, A070826, A143293, A203008, A250130, A327860, A348301, A369651.
Cf. A053144 (a lower bound), A074107 (an upper bound).
Cf. A109628 (indices k where a(k) is prime), A244622 (corresponding primes), A244621 (a(n) mod 12).
Cf. A369972 (k where prime(1+k)|a(k)), A369973 (corresponding primorials), A293457 (corresponding primes), A377992 (antiderivatives of the terms > 1 of this sequence).
Cf. also A223037, A260615, A274070, A327978, A353299, A353534, A356253.

[ Numerator(&+[ NthPrime(k)^-1: k in [1..n]]): n in [1..18] ];  // Bruno Berselli, Apr 11 2011

h:= n-> add(1/(ithprime(i)),i=1..n);
t1:=[seq(h(n),n=0..50)];
t1a:=map(numer,t1); # A024451
t1b:=map(denom,t1); # A002110 - N. J. A. Sloane, Apr 25 2014

a[n_] := Numerator @ Sum[1/Prime[i], {i, n}]; Array[a,18]  (* Jean-François Alcover, Apr 11 2011 *)
f[k_] := Prime[k]; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 16}] (* A024451 *)
(* Clark Kimberling, Dec 29 2011 *)
Numerator[Accumulate[1/Prime[Range[20]]]] (* Harvey P. Dale, Apr 11 2012 *)

a(n) = numerator(sum(i=1, n, 1/prime(i))); \\ Michel Marcus, Sep 18 2018

from sympy import prime
from fractions import Fraction
def a(n): return sum(Fraction(1, prime(k)) for k in range(1, n+1)).numerator
print([a(n) for n in range(20)]) # Michael S. Branicky, Feb 12 2021

from math import prod
from sympy import prime
def A024451(n):
    q = prod(plist:=tuple(prime(i) for i in range(1,n+1)))
    return sum(q//p for p in plist) # Chai Wah Wu, Nov 03 2022

Limit_{n->oo} (Sum_{p <= n} 1/p - log log n) = 0.2614972... = A077761.
a(n) = (Product_{i=1..n} prime(i))*(Sum_{i=1..n} 1/prime(i)). - Benoit Cloitre, Jan 30 2002
(n+1)-st elementary symmetric function of the first n primes.
a(n) = a(n-1)*A000040(n) + A002110(n-1). - Henry Bottomley, Sep 27 2006
From Antti Karttunen, Jan 31 2024, Feb 08 2024 and Nov 19 2024: (Start)
a(0) = 0, for n > 0, a(n) = 2*A203008(n-1) + A070826(n).
For n > 0, a(n) = A327860(A143293(n-1)).
For n > 0, a(n) = A348301(n) + A002110(n).
For n = 3..175, a(n) = A356253(A002110(n)). [See comments in A356253.]
For n >= 0, A053144(n) <= a(n) <= A074107(n) < A070826(1+n).
(End)

a(0)=0 prepended by Alois P. Heinz, Jun 26 2015

A014701 Number of multiplications to compute n-th power by the Chandah-sutra method.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 5, 6, 6, 7, 6, 7, 7, 8, 6, 7, 7, 8, 7, 8, 8, 9, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 7, 8, 8, 9, 8, 9, 9

Offset: 1

James Kilfiger (jamesk(AT)maths.warwick.ac.uk)

In other words, number of steps to reach 1 starting from n and using the process: x -> x-1 if n is odd and x -> x/2 otherwise.
a(n) = number of 0's + twice number of 1's (disregarding the leading digit 1) in the binary expansion of n, i.e., A007088(n). - Lekraj Beedassy, May 28 2010
From Daniel Forgues, Jul 31 2012: (Start)
For the binary Fibonacci rabbits sequence (A036299) (cf. OEIS Wiki link below) we have the substitution/concatenation rule: a(n), n >= 3, may be obtained by the concatenation of a(n-1) and a(n-2), with a(1) = 0, a(2) = 1. Thus, using . (dot) as the concatenation operator, we have the recursive substitution/concatenation
    a(n) = a(n-0)
    a(n) = a(n-1).a(n-2)
    a(n) = a(n-2).a(n-3).a(n-3).a(n-4)
    a(n) = a(n-3).a(n-4).a(n-4).a(n-5).a(n-4).a(n-5).a(n-5).a(n-6)
  which suggests the sequence
    {0}
    {1, 2}
    {2, 3, 3, 4}
    {3, 4, 4, 5, 4, 5, 5, 6}
  whose concatenation gives A014701 (this sequence).
Number of multiplications to compute n-th power by the Chandah-sutra method, also called left-to-right binary exponentiation:
  x^1 = x^(   1_2) =                                 (x)   (0 prod)
  x^2 = x^(  10_2) =                         (x^2)         (1 prod)
  x^3 = x^(  11_2) =                         (x^2) * (x)   (2 prod)
  x^4 = x^( 100_2) =               (x^2)^2                 (2 prod)
  x^5 = x^( 101_2) =               (x^2)^2         * (x)   (3 prod)
  x^6 = x^( 110_2) =               (x^2)^2 * (x^2)         (3 prod)
  x^7 = x^( 111_2) =               (x^2)^2 * (x^2) * (x)   (4 prod)
  x^8 = x^(1000_2) = ((x^2)^2)^2                           (3 prod) (End)
From Ya-Ping Lu, Mar 03 2021: (Start)
Index at which record m occurs is A052955(m).
First appearance of m in the sequence (or the record value m) is at n = 2^(m/2 + 1) - 1 for even m, and at n = 3*2^((m - 1)/2) - 1 for odd m.
The last appearance of m in the sequence is at n = 2^m. (End)
a(n) is the digit sum of n-1 in bijective base-2. Since the Fibonacci number F(m) can be defined as the number of ways to compose m as the sum of 1s and 2s, we get that m appears F(m) times in the sequence. - Oscar Cunningham, Apr 14 2024
Conjecture: a(n+1) is the minimal number of steps to go from 0 to n, by choosing before each step, after the first step, whether to keep the same step length or double it. The initial step length is 1. - Jean-Marc Rebert, May 15 2025

			5 -> 4 -> 2 -> 1 so 3 steps are needed to reach 1 hence a(5)=3; 9 -> 8 -> 4 -> 2 -> 1 hence a(9)=4.

Alois P. Heinz, Table of n, a(n) for n = 1..16384 (first 1000 terms from T. D. Noe)
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints (2025).
C. K. Caldwell, The Prime Glossary, binary exponentiation
Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length, Proceedings of the 46th International Symposium on Mathematical Foundations of Computer Science, Article No. 53, pp. 53:1-53:15, 2021.
J. Jordan and R. Southwell, Further Properties of Reproducing Graphs, Applied Mathematics, Vol. 1 No. 5, 2010, pp. 344-350. - From _N. J. A. Sloane_, Feb 03 2013
Szymon Łukaszyk and Wawrzyniec Bieniawski, Assembly Theory of Binary Messages (How to Assemble a Black Hole and Use it to Assemble New Binary Information?), Preprints (2024).
OEIS Wiki, Binary Fibonacci rabbits sequence
Index to sequences related to the complexity of n
Index entries for sequences related to binary expansion of n

Cf. A003313, A056791, A056792, A115964, A052955.

a014701 1 = 0
a014701 n = a007953 $ a007931 (n - 1)
-- Reinhard Zumkeller, Oct 26 2012

A014701 := proc(n) local j,k; j := n; k := 0; while(j>1) do if j mod 2=1 then j := j-1 else j := j/2 fi; k := k+1 od end;
# second Maple program:
a:= n-> add(i+1, i=Bits[Split](n))-2:
seq(a(n), n=1..128);  # Alois P. Heinz, Aug 30 2021

a[n_] := DigitCount[n, 2] /. {x_, y_} -> 2x + y - 2; Array[a, 100] (* Robert G. Wilson v, Jul 31 2012 *)

a(n)=hammingweight(n)+logint(n,2)-1 \\ Charles R Greathouse IV, Dec 29 2016

def a(n):
    if n==1:
        return 0
    return a(n//2)+1+n%2
for i in range(1,60):
    print(a(i), end=", ")
# Pablo Hueso Merino, Oct 28 2020

a(n) = A056792(n) - 1 = A056791(n) - 2.
a(n) = floor(log_2(n)) + (number of 1's in binary representation of n) - 1. - Corrected (- 1 at end) by Daniel Forgues, Aug 01 2012
a(2^n) = n, a(2^n-1) = 2*(n-1), and for n >= 2, log_2(n) <= a(n) <= 2*log_2(n) - 1. - Robert FERREOL, Oct 01 2014
Let u(1) = 1, u(2*n) = u(n)+1, u(2*n+1) = u(2*n)+1; then a(1) = 0 and a(n) = u(n-1). - Benoit Cloitre, Dec 19 2002
G.f.: -2/(1-x) + (1/(1-x)) * Sum_{k>=0} (2*x^2^k + x^2^(k+1))/(1+x^2^k). - Ralf Stephan, Aug 15 2003
From {0}, apply the substitution rule (n -> n+1, n+2) repeatedly, giving {{0}, {1, 2}, {2, 3, 3, 4}, {3, 4, 4, 5, 4, 5, 5, 6}, ...} and concatenate. - Daniel Forgues, Jul 31 2012
For n > 1: a(n) = A007953(A007931(n-1)). - Reinhard Zumkeller, Oct 26 2012
a(n) >= A003313(n). - Charles R Greathouse IV, Jan 03 2018
a(n) = a(floor(n/2)) + 1 + (n mod 2) for n > 1. - Pablo Hueso Merino, Oct 28 2020
a(n+1) = max_{1<=i<=n} (H(i) + H(n-i)) where H(n) denotes the Hamming weight of n (A000120(n)). See Lemma 8 in Gruber/Holzer 2021 article. - Hermann Gruber, Jun 26 2024

A372695 Cubefull numbers that are not prime powers.

Original entry on oeis.org

216, 432, 648, 864, 1000, 1296, 1728, 1944, 2000, 2592, 2744, 3375, 3456, 3888, 4000, 5000, 5184, 5488, 5832, 6912, 7776, 8000, 9261, 10000, 10125, 10368, 10648, 10976, 11664, 13824, 15552, 16000, 16875, 17496, 17576, 19208, 20000, 20736, 21296, 21952, 23328, 25000

Offset: 1

Michael De Vlieger, May 14 2024

nonn

Numbers k such that rad(k)^3 | k and omega(k) > 1. In other words, numbers with at least 2 distinct prime factors whose prime power factors have exponents that exceed 2.
Proper subset of the following sequences: A001694, A036966, A126706, A286708.
Superset of A372841.
Smallest term k with omega(k) = m is k = A002110(m)^3 = A115964(m).

			Table of smallest 12 terms and instances of omega(a(n)) = m for m = 2..4
    n      a(n)
  ------------------------
    1      216 = 2^3 * 3^3
    2      432 = 2^4 * 3^3
    3      648 = 2^3 * 3^4
    4      864 = 2^5 * 3^3
    5     1000 = 2^3 * 5^3
    6     1296 = 2^4 * 3^4
    7     1728 = 2^6 * 3^3
    8     1944 = 2^3 * 3^5
    9     2000 = 2^4 * 5^3
   10     2592 = 2^5 * 3^4
   11     2744 = 2^3 * 7^3
   12     3375 = 3^3 * 5^3
  ...
   43    27000 = 2^3 * 3^3 * 5^3
  ...
  587  9261000 = 2^3 * 3^3 * 5^3 * 7^3

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A001694, A007947, A024619, A036966, A115964, A126706, A286708, A372841.

Cf. A065483, A152441.

nn = 25000; Rest@ Select[Union@ Flatten@ Table[a^5 * b^4 * c^3, {c, Surd[nn, 3]}, {b, Surd[nn/(c^3), 4]}, {a, Surd[nn/(b^4 * c^3), 5]}], Not@*PrimePowerQ]

from math import gcd
from sympy import primepi, integer_nthroot, factorint
def A372695(n):
    def f(x):
        c = n+1+x+sum(primepi(integer_nthroot(x, k)[0]) for k in range(3, x.bit_length()))
        for w in range(1,integer_nthroot(x,5)[0]+1):
            if all(d<=1 for d in factorint(w).values()):
                for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                    if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c -= integer_nthroot(z//y**4,3)[0]
        return c
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    return bisection(f,n,n) # Chai Wah Wu, Sep 12 2024

Intersection of A036966 and A024619.

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/(p^2*(p-1))) - Sum_{p prime} 1/(p^2*(p-1)) - 1 = A065483 - A152441 - 1 = 0.0188749045... . - Amiram Eldar, May 17 2024

A115963 Numerator of Sum_{i=1..n} 1/prime(i)^3.

Original entry on oeis.org

1, 35, 4591, 1601713, 2141141003, 4716413174591, 23198819007792583, 159253748925534977797, 1938552948676080555065099, 47290471293028435532185602511, 1409101231790431848106470385672201

Offset: 1

Jonathan Vos Post, Mar 14 2006

Denominators = A115964. See also: A024451 Numerator of Sum_{i=1..n} 1/prime(i). A002110 Primorial [denominator of Numerator of Sum_{i=1..n} 1/prime(i)]. A061015 Numerator of Sum_{i=1..n} 1/prime(i)^2.

			1/8, 35/216, 4591/27000, 1601713/9261000, 2141141003/12326391000, 4716413174591/27081081027000.

Cf. A000040, A075986/A075987, A106830/A034386.

Accumulate[1/Prime[Range[20]]^3]//Numerator (* Harvey P. Dale, Dec 30 2024 *)

a(n) = Numerator of Sum_{i=1..n} 1/A000040(i)^3.

A241189 Numerator of Sum_{i=1..n} 1/(prime(i)*prime(i+1)).

Original entry on oeis.org

1, 7, 11, 127, 1693, 29243, 561623, 13019431, 379503437, 11809225121, 438235268123, 18007758091069, 775817745542929, 36524284093223105, 1938403609207158571, 2160165866032831207, 131893095784520401909, 8844093116997411126541, 628373208972323386101329, 45900898298568589325230523

Offset: 1

N. J. A. Sloane, Apr 25 2014, based on a suggestion from Timothy Varghese

a(371) has 1002 decimal digits. - Michael De Vlieger, Jan 27 2016

			1/6, 7/30, 11/42, 127/462, 1693/6006, 29243/102102, 561623/1939938, 13019431/44618574, 379503437/1293938646, 11809225121/40112098026, 438235268123/1484147626962, ...

Michael De Vlieger, Table of n, a(n) for n = 1..370

Cf. A024451/A002110, A241189/A241190, A241191/A241192.

See also A061015/A061742, A115963/A115964.

g:= n-> add(1/(ithprime(i)*ithprime(i+1)),i=1..n);
t1:=[seq(g(n),n=1..20)];
t1a:=map(numer,t1); # A241189
t1b:=map(denom,t1); # A241190

Table[Numerator@ Sum[1/(Prime[i + 1] Prime@ i), {i, n}], {n, 20}] (* Michael De Vlieger, Jan 27 2016 *)
Accumulate[1/#&/@(Times@@@Partition[Prime[Range[25]],2,1])]//Numerator (* Harvey P. Dale, Mar 14 2023 *)

A304117 If n = Product (p_j^k_j) then a(n) = Product (pi(p_j)*k_j), where pi() = A000720.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 4, 3, 4, 3, 5, 4, 6, 4, 6, 4, 7, 4, 8, 6, 8, 5, 9, 6, 6, 6, 6, 8, 10, 6, 11, 5, 10, 7, 12, 8, 12, 8, 12, 9, 13, 8, 14, 10, 12, 9, 15, 8, 8, 6, 14, 12, 16, 6, 15, 12, 16, 10, 17, 12, 18, 11, 16, 6, 18, 10, 19, 14, 18, 12, 20, 12, 21, 12, 12, 16, 20, 12, 22, 12

Offset: 1

Ilya Gutkovskiy, May 06 2018

			a(36) = 8 because 36 = 2^2*3^2 = prime(1)^2*prime(2)^2 and 1*2*2*2 = 8.

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Ilya Gutkovskiy, Extended graphical example
Index entries for sequences computed from indices in prime factorization
Index entries for sequences computed from exponents in factorization of n

Cf. A000026, A000040, A000720, A001414, A002110, A003963, A005361, A056239, A156061, A304037.

a[n_] := Times @@ (PrimePi[#[[1]]] #[[2]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 1, 80}]

a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = primepi(f[k,1])*f[k,2]; f[k, 2] = 1); factorback(f); \\ Michel Marcus, May 06 2018

a(n) = A005361(n)*A156061(n).
a(p^k) = A000720(p)*k where p is a prime.
a(A002110(m)^k) = k^m*m!.
As an example:
a(A000040(k)) = k.
a(A006450(k)) = A000040(k).
a(A001248(k)) = a(A031215(k)) = A005843(k).
a(A030078(k)) = a(A031336(k)) = A008585(k)
a(A061742(k)) = A000165(k).
a(A115964(k)) = A032031(k).
a(A002110(k)) = A000142(k).
a(A080696(k)) = A002110(k).

A304037 If n = Product (p_j^k_j) then a(n) = Sum (pi(p_j)^k_j), where pi() = A000720.

Original entry on oeis.org

0, 1, 2, 1, 3, 3, 4, 1, 4, 4, 5, 3, 6, 5, 5, 1, 7, 5, 8, 4, 6, 6, 9, 3, 9, 7, 8, 5, 10, 6, 11, 1, 7, 8, 7, 5, 12, 9, 8, 4, 13, 7, 14, 6, 7, 10, 15, 3, 16, 10, 9, 7, 16, 9, 8, 5, 10, 11, 17, 6, 18, 12, 8, 1, 9, 8, 19, 8, 11, 8, 20, 5, 21, 13, 11, 9, 9, 9, 22, 4, 16, 14, 23, 7, 10, 15, 12, 6

Offset: 1

Ilya Gutkovskiy, May 05 2018

nonn

			a(72) = 5 because 72 = 2^3*3^2 = prime(1)^3*prime(2)^2 and 1^3 + 2^2 = 5.

Ilya Gutkovskiy, Extended graphical example
Index entries for sequences computed from indices in prime factorization

Cf. A000040, A000720, A002110, A003963, A008475, A056239, A066328, A156061, A222416.

a[n_] := Plus @@ (PrimePi[#[[1]]]^#[[2]]& /@ FactorInteger[n]); a[1] = 0; Table[a[n], {n, 1, 88}]

If gcd(u,v) = 1 then a(u*v) = a(u) + a(v).
a(p^k) = A000720(p)^k where p is a prime.
a(A002110(m)^k) = 1^k + 2^k + ... + m^k.
As an example:
a(A000040(k)) = k.
a(A006450(k)) = A000040(k).
a(A038580(k)) = A006450(k).
a(A001248(k)) = a(A011757(k)) = A000290(k).
a(A030078(k)) = a(A055875(k)) = A000578(k).
a(A002110(k)) = a(A011756(k)) = A000217(k).
a(A061742(k)) = A000330(k).
a(A115964(k)) = A000537(k).
a(A080696(k)) = A007504(k).
a(A076954(k)) = A001923(k).

A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).

Original entry on oeis.org

1, 2, 3, 4, 3, 2, 3, 8, 9, 10, 3, 12, 3, 10, 3, 16, 3, 18, 3, 20, 21, 10, 3, 24, 9, 10, 27, 20, 3, 2, 3, 32, 21, 10, 3, 4, 3, 10, 21, 40, 3, 10, 3, 20, 45, 10, 3, 48, 9, 50, 21, 20, 3, 54, 21, 40, 21, 10, 3, 12, 3, 10, 63, 64, 21, 10, 3, 20, 21, 10, 3, 72, 3

Offset: 1

Rémy Sigrist, Jul 23 2022

nonn

We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.

This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.

			For n = 99:
- 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
- the list of exponents is: 1 0 0 2 0,
- compressing consecutive values, we obtain: 1 0 2 0,
- so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.

Cf. A007814, A061742, A067255, A090079, A115964, A294674, A319521, A337864, A356008, A356021 (fixed points).

a(n) = { my (v=1, e=-1, k=0); forprime (p=2, oo, if (n==1, return (v), if (e!=e=valuation(n,p), v*=prime(k++)^e); n/=p^e)) }

a(a(n)) = a(n).
a(n^k) = a(n)^k for any k >= 0.
a(n) = A319521(A356008(n)).
A007814(a(n)) = A007814(n).
a(n) = 3 iff n belongs to A294674 \ {1}.
a(n) = 4 iff n belongs to A061742 \ {1}.
a(n) = 8 iff n belongs to A115964.

A361810 a(n) is the sum of divisors of n that are both infinitary and exponential.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 10, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 30, 25, 26, 30, 28, 29, 30, 31, 34, 33, 34, 35, 36, 37, 38, 39, 50, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 60, 55, 70, 57, 58, 59, 60, 61, 62, 63, 68, 65, 66, 67, 68

Offset: 1

Amiram Eldar, Mar 25 2023

The number of these divisors is A359411(n).

The indices of records of a(n)/n are the primorials (A002110) cubed, i.e., 1 and the terms of A115964.

			a(8) = 10 since 8 has 2 divisors that are both infinitary and exponential, 2 and 8, and 2 + 8 = 10.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A002110, A049417, A051377, A082020, A115964, A138302, A359411.

f[p_, e_] := DivisorSum[e, p^# &, BitOr[#, e] == e &]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

s(p,e) = sumdiv(e, d, p^d*(bitor(d, e) == e));
a(n) = {my(f = factor(n)); prod(i = 1, #f~, s(f[i, 1], f[i, 2])); }

Multiplicative with a(p^e) = Sum_{d|e, bitor(d, e) == e} p^d.
a(n) >= n, with equality if and only if n is in A138302.
limsup_{n->oo} a(n)/n = Product_{p prime} (1 + 1/p^2) = 15/Pi^2 (A082020).
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{p prime} ((1 - 1/p)*(1 + Sum_{e>=1} Sum_{d|e, bitor(d, e) == e} p^(d-2*e))) = 0.51015879911178031024... .

A359412 Numbers with a record number of divisors that are both infinitary and exponential.

Original entry on oeis.org

1, 8, 216, 27000, 9261000, 12326391000, 27081081027000, 110924107886592000, 544970142046826496000, 3737950204299182936064000, 45479640135708158783090688000, 1109202943269786284560798789632000, 33044264882950203203350756741926912000, 1673791149116076642859325881248823873536000

Offset: 1

Amiram Eldar, Dec 30 2022

nonn

Indices of records in A359411.
a(2)-a(7) are the first 6 terms of A115964.
The first 15 terms are cubes. Are there noncubes in this sequence?
The corresponding record values are 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, ... . Apparently, this sequence of records is the powers of 2 (A000079).

Amiram Eldar, Table of n, a(n) for n = 1..15

Cf. A000079, A000578, A115964, A359411.
Subsequence of A025487.
Similar sequences: A037992, A318278.

s[n_] := DivisorSum[n, 1 &, BitAnd[n, #] == # &]; f[p_, e_] := s[e]; d[1] = 1; d[n_] := Times @@ f @@@ FactorInteger[n];
v = Cases[Import["https://oeis.org/A025487/b025487.txt", "Table"], {, }][[;; , 2]];
seq = {}; dm = 0; Do[If[(dk = d[v[[k]]]) > dm, dm = dk; AppendTo[seq, v[[k]]]], {k, 1, Length[v]}]; seq

cp's OEIS Frontend

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