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.

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A240905 Smallest k such that the minimal factor in factorization of k! over distinct terms of A050376 is A050376(n), or a(n) = 0 if there is no such k.

Original entry on oeis.org

2, 12, 20, 6, 10, 130, 180, 240, 480, 597, 901, 40537, 15841, 23401, 36720, 112321, 20377, 177842, 101755, 855369, 2333282, 654549, 15078026, 9206403, 5829606, 75690506, 64946036, 650025768, 1100265039, 1966444000, 6660851430, 1505032794, 7305026242, 42063043872, 54868098259
Offset: 1

Views

Author

Vladimir Shevelev, Apr 14 2014

Keywords

Comments

a(n) is the smallest k such that the minimal infinitary divisor of k! is A050376(n).
Conjecture: a(n) > 0 for all n.
a(28) > 2.5*10^8, if it exists. - Amiram Eldar, Jun 18 2025
a(34) > 2.8*10^10 if it exists. - David A. Corneth, Jun 19 2025

Examples

			Let n = 4. A050376(4)=5. For k = 2, 3, 4, 5, 6, we have the following factorizations over distinct terms of A050376: 2! = 2, 3! = 2*3, 4! = 2*3*4, 5! = 2*3*4*5, 6! = 5*9*16. Only the last factorization begins with 5. So a(4) = 6.
From _David A. Corneth_, Jun 19 2025: (Start)
a(6) = 130. Once we checked that a(6) is > 125 we try 126. The minimal factor of k! into distinct products must be A050376(6) = 8. For 126! we have the 5-adic valuation of 31 so the minimal factor is at most 5.
To get rid of the 5 we try the next candidate > 126 that is a multiple of 5. This is 130. We can just skip 127, 128 and 129 altogether. It turns out this smallest factor for 130! is 8 giving the value for a(6). (End)
		

Crossrefs

Programs

  • PARI
    \\ See Corneth link
    
  • PARI
    \\ See Eldar link

Extensions

More terms from Peter J. C. Moses, Apr 19 2014
a(19)-a(27) from Amiram Eldar, Jun 18 2025
a(28)-a(35) from David A. Corneth, Jun 18, Jun 21 2025

A330687 Positions of records in A050377, number of ways to factor n into "Fermi-Dirac primes" (A050376).

Original entry on oeis.org

1, 4, 16, 64, 144, 256, 576, 1024, 1296, 2304, 5184, 9216, 20736, 82944, 186624, 331776, 746496, 1327104, 2073600, 2985984, 5308416, 8294400, 18662400, 21233664, 26873856, 33177600, 47775744, 51840000, 74649600, 107495424, 132710400, 207360000, 429981696, 530841600, 671846400, 829440000, 1194393600, 1719926784, 1866240000, 2687385600
Offset: 1

Views

Author

Antti Karttunen, Dec 28 2019

Keywords

Comments

From David A. Corneth, Dec 29 2019: (Start)
Each term is a perfect square. Proof: A050377(n) is multiplicative with a(p^e) = A018819(e) and A018819(2k) = A018819(2k+1) and this sequence considers just records so we only need exponents of the form 2k; i.e., terms are squares.
Furthermore, the exponent 2 occurs at most once in the prime factorization of a(n) as A018819(2)^2 = A018819(4) = 4. So if the last two exponents in the prime factorization of m are 2's then setting the first of those two exponents to 4 and the other to 0 gives the same A050377(m).
Example of an application of this proof: we have 3600 = 2^4 * 3^2 * 5^2. We see the last two exponents are 2's so we can set the first of those two to 4 and the second to 0. This gives 2^4 * 3^4 = 1296 and, indeed, A050377(1296) = A050377(3600) = 16.
It seems that most exponents of a(n) are divisible by 4.
More specifically: Let S(n) be the list, possibly with duplicates, of exponents occurring in the prime factorizations of terms with the sum of exponents in the prime factorization <= n.
Let R(n) = |{x : x==4, S(n)}| / |S(n)|.
For example, S(8) is found from the following terms: 4, 16, 64, 144, 256, 576 and 1296 as the exponents in the prime factorization are (2), (4), (6), (4, 2), (8), (6, 2), (4, 4). The sums of each of these exponents per term is <= 8. There are 10 exponents listed. Of these 10 there are 5 that are divisible by 4. Therefore R(8) = 5/10.
Then it seems that R(n) tends to some value > 0.8 as n grows. (End)

Crossrefs

Cf. A018819, A050376, A050377, A329900, A330684 (square roots), A330688 (the record values), A330689 (primorial deflation).
Subsequence of A025487.

Programs

  • Mathematica
    Block[{s = Rest@ Nest[Function[{a, n, b}, Append[a, {Times @@ Map[a[[# + 1, -1]] &, FactorInteger[n][[All, -1]] ], b}]] @@ {#1, #2, #1[[-1, -1]] + If[EvenQ@ #2, #1[[#2/2 + 1, -1]], 0 ]} & @@ {#, Length@ #} &, {{0, 1}, {1, 1}}, 10^5][[All, 1]], t}, t = Union@ FoldList[Max, s]; Map[FirstPosition[s, #][[1]] &, t]] (* Michael De Vlieger, Dec 29 2019 *)
  • PARI
    upto_e = 101; \\ 101 --> 211 terms.
    A018819(n) = if( n<1, n==0, if( n%2, A018819(n-1), A018819(n/2)+A018819(n-1))); \\ From A018819
    v018819 = vector(upto_e,n,A018819(n)); \\ Precompute.
    A050377(n) = factorback(apply(e -> v018819[e], factor(n)[, 2]));
    A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)}; \\ From A283980
    A330687list(e) = { my(lista = List([1, 2]), i=2, u = 2^e, t, m=0, v025487); while(lista[i] != u, if(2*lista[i] <= u, listput(lista,2*lista[i]); t = A283980(lista[i]); if(t <= u, listput(lista,t))); i++); v025487 = vecsort(Vec(lista)); lista = List([]); for(i=1,#v025487,if((t=A050377(v025487[i]))>m, listput(lista,v025487[i]); m=t)); Vec(lista); };
    v330687 = A330687list(upto_e);
    A330687(n) = v330687[n];
    for(n=1,#v330687,print1(A330687(n),", "));

Formula

A050377(a(n)) = A330688(n).
A329900(a(n)) = A330689(n).
a(n) = A330684(n)^2.

A177334 Largest factor in the factorization of n! over distinct terms of A050376.

Original entry on oeis.org

2, 3, 4, 5, 16, 16, 16, 81, 256, 256, 256, 256, 256, 256, 256, 256, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 4294967296, 4294967296, 4294967296, 4294967296, 4294967296, 4294967296
Offset: 2

Views

Author

Vladimir Shevelev, May 06 2010

Keywords

Comments

Each number >=2 has a unique factorization over distinct terms of A050376.
This is obtained from the standard prime factor representation by splitting the exponents into a sum of powers of 2, and further factorization according to the nonzero term of this base-2 representation.
The largest factor of this representation of A000142(n) defines this sequence.

References

  • V. S. Shevelev, Multiplicative functions in the Fermi-Dirac arithmetic, Izvestia Vuzov of the North-Caucasus region, Nature sciences 4 (1996), 28-43 [Russian].

Crossrefs

Programs

  • Maple
    A177334 := proc(n) local a,p,pow2 ; a := 1 ; for p in ifactors(n!)[2] do pow2 := convert( op(2,p),base,2) ; for j from 1 to nops(pow2) do if op(j,pow2) <> 0 then a := max(a,op(1,p)^(2^(j-1))) ; end if; end do: end do: return a ; end proc:
    seq(A177334(n),n=2..60) ; # R. J. Mathar, Jun 16 2010
  • Mathematica
    b[n_] :=2^(-1+Position[ Reverse@IntegerDigits[n, 2],?(#==1&)])//Flatten; a[n] := Module[{np = PrimePi[n]}, v=Table[0,{np}]; Do[p = Prime[k]; Do[v[[k]] += IntegerExponent[j, p], {j,2,n}],  {k,1,np}]; Max[(Prime/@Range[np])^(b/@v) // Flatten]]; Array[a, 38, 2]  (* Amiram Eldar, Sep 17 2019 *)

Extensions

a(18) and a(19) corrected and sequence extended by R. J. Mathar, Jun 16 2010

A182979 Fermi-Dirac representation of n. Let n have factorization p1^(2^e1) * p2^(2^e2) * ... * pr^(2^er), where each factor is in A050376. The number n is represented by a binary string that indicates which terms of A050376 appear in the factorization of n.

Original entry on oeis.org

0, 1, 10, 100, 1000, 11, 10000, 101, 100000, 1001, 1000000, 110, 10000000, 10001, 1010, 100000000, 1000000000, 100001, 10000000000, 1100, 10010, 1000001, 100000000000, 111, 1000000000000, 10000001, 100010, 10100, 10000000000000, 1011, 100000000000000, 100000001, 1000010, 1000000001, 11000, 100100
Offset: 1

Views

Author

Daniel Forgues, Feb 10 2011, Feb 13 2011

Keywords

Comments

Every number has a unique representation as a product of terms from A050376. - N. J. A. Sloane, Feb 11 2011
The "Fermi-Dirac factorization" of n, i.e., the factorization of n into prime powers of the form p_k^(2^e_k), e_k >= 0, (A050376) allows each of those prime powers to be used at most once, since this corresponds to the binary representation of the exponents of the prime powers p^a of the "Bose-Einstein factorization" of n, i.e., the classic prime factorization of n. (Cf. A050376 comments.)
The prime powers of the form p_k^(2^e_k), e_k >= 0 (A050376) might be called "Fermi-Dirac primes" since they may appear at most once (thus raised to powers 0 or 1) in the "Fermi-Dirac factorization" of n. Compare with the classic prime factorization of n, which might be called the "Bose-Einstein factorization" of n, where the primes (which might be called "Bose-Einstein primes") may appear any number of times >= 0.
In the "Fermi-Dirac representation" of n, if a given prime power with powers of two as exponents does not appear in the factorization of n into prime powers with powers of two as exponents, we use 0 as a placeholder; otherwise, we use 1 to indicate that the given prime power with powers of two as exponents does appear in the "Fermi-Dirac factorization" of n.
In the base-b representation of n, we do not show the leading 0's, except for 0 where it is more convenient to show it than to show nothing. Similarly, for the "Fermi-Dirac representation" of n, we do not show the leading 0's, except for 0, which is the representation of 1, where it is more convenient to show it than to show nothing.
The limit of the supremum of the number of "binary digits" of the representation of n is asymptotic to the number of primes up to n, i.e., n/log(n), making this representation absolutely impractical!
See A052330 for the numbers having representation as 0, 1, 10, 11, 100, 101, 110, 111, ... which is an ordering of the positive integers. (Cf. OEIS Wiki page.)
Let n have factorization (f_r)^g_r * ... * (f_2)^g_2 * (f_1)^g_1, where f_i is the i-th prime power of the form p_k^(2^e_k), e_k >= 0 (A050376, A302778); then a(n) = Sum_{i=1..r} g_i * 2^(i-1).
The number of 1's in a(n) is the number of terms of A050376 dividing n with odd maximal exponent. For example, if n=96, then the maximal exponent of 2 that divides 96 is 5, for 3 it is 1, for 4 it is 2, for 16 it is 1. Thus only 2, 3 and 16 divide n with odd maximal exponents. Therefore, the number of 1's in a(96) is 3. Moreover, since 2=A050376(1), 3=A050376(2) and 16=A050376(9), then 1's appear in positions 1,2,9 from the right. - Vladimir Shevelev, Nov 02 2013

Examples

			"Fermi-Dirac factorizations" (cf. A050376 examples, here with exponents of "Fermi-Dirac primes" being either 0 or 1):
6 = 3*2 = 3^1*2^1, so a(6) = 11;
8 = 4*2 = 4^1*3^0*2^1, so a(8) = 101;
20 = 5*4 = 5^1*4^1*3^0*2^0, so a(20) = 1100;
24 = 4*3*2 = 4^1*3^1*2^1, so a(24) = 111;
27 = 9*3 = 9^1*7^0*5^0*4^0*3^1*2^0, so a(27) = 100010;
32 = 16*2 = 16^1*13^0*11^0*9^0*7^0*5^0*4^0*3^0*2^1, so a(32) = 100000001;
64 = 16*4 = 16^1*13^0*11^0*9^0*7^0*5^0*4^1*3^0*2^0, so a(64) = 100000100;
108 = 9*4*3 = 9^1*7^0*5^0*4^1*3^1*2^0, so a(108) = 100110;
120 = 5*4*3*2 = 5^1*4^1*3^1*2^1, so a(120) = 1111;
...
		

Crossrefs

Programs

  • Mathematica
    nn=24; p=Select[Range[nn], PrimeQ]; Do[p=Select[Union[p,p^2], #<=nn&], {Floor[Log[2,Log[2,nn]]]}]; Table[m=n; FromDigits[Table[If[Mod[m,i]==0, m=m/i; 1, 0], {i,Reverse[p]}]],{n,nn}]

Formula

Let q_1,q_2,q_3,... be consecutive terms of A050376 and n = q_1^a_1 * q_2^a_2 *...* q_r^a_r, where a_i = 0 or 1. Then a(n) = a_1 + 10*a_2 + ... +10^(r-1)*a_r. For example, since 30 = 2^1 * 3^1 * 4^0 * 5^1, then a(30)= 1 + 10 + 1000 = 1011. - Vladimir Shevelev, Nov 02 2013
a(n) = A007088(A052331(n)). - Antti Karttunen, Apr 17 2018

Extensions

Clearer definition from T. D. Noe, Feb 11 2011
Edited by N. J. A. Sloane, Jul 21 2018

A302778 Number of "Fermi-Dirac primes" (A050376) <= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 8, 9, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 22, 22, 23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 25, 25, 26, 26, 26, 26, 26, 26, 27, 27, 28, 28, 29, 29, 29, 29, 29, 29, 30
Offset: 1

Views

Author

Antti Karttunen, Apr 16 2018

Keywords

Crossrefs

Partial sums of A302777. A left inverse of A050376.
Differs from A203967 for the first time at n=64, where a(64) = 23, while A203967(64) = 24.
Cf. also A000720, A025528.

Programs

  • Mathematica
    s[n_] := Boole[n > 1 && Length[(f = FactorInteger[n])] == 1 && (e = f[[;; , 2]]) == 2^IntegerExponent[e, 2]]; Accumulate @ Array[s, 100] (* Amiram Eldar, Nov 27 2020 *)
  • PARI
    A209229(n) = (n && !bitand(n,n-1));
    A302777(n) = A209229(isprimepower(n));
    s=0; for(n=1,105,s+=A302777(n); print1(s,", "));
    
  • Python
    from sympy import primepi, integer_nthroot
    def A302778(n): return sum(primepi(integer_nthroot(n,1<Chai Wah Wu, Feb 18-19 2025

Formula

a(1) = 0; for n > 1, a(n) = A302777(n) + a(n-1).
For all n >= 1, a(A050376(n)) = n.

A319826 GCD of the strict integer partition with FDH number n; GCD of the indices (in A050376) of Fermi-Dirac prime factors of n.

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 2, 9, 10, 1, 11, 1, 1, 1, 12, 1, 13, 1, 2, 1, 14, 1, 15, 1, 1, 1, 1, 3, 16, 1, 2, 1, 17, 1, 18, 1, 2, 1, 19, 1, 20, 1, 2, 1, 21, 1, 1, 1, 1, 1, 22, 1, 23, 1, 1, 3, 4, 1, 24, 1, 2, 1, 25, 1, 26, 1, 1, 1, 1, 1, 27, 1, 28
Offset: 1

Views

Author

Gus Wiseman, Sep 28 2018

Keywords

Comments

Let f(n) = A050376(n) be the n-th Fermi-Dirac prime. The FDH number of a strict integer partition (y_1, ..., y_k) is f(y_1) * ... * f(y_k).

Examples

			45 is the FDH number of (6,4), which has GCD 2, so a(45) = 2.
		

Crossrefs

Programs

  • Mathematica
    nn=200;
    FDfactor[n_]:=If[n==1,{},Sort[Join@@Cases[FactorInteger[n],{p_,k_}:>Power[p,Cases[Position[IntegerDigits[k,2]//Reverse,1],{m_}->2^(m-1)]]]]];
    FDprimeList=Array[FDfactor,nn,1,Union];FDrules=MapIndexed[(#1->#2[[1]])&,FDprimeList];
    GCD@@@Table[Reverse[FDfactor[n]/.FDrules],{n,nn}]
  • PARI
    A319826(n) = { my(i=1,g=0,x=A052331(n)); while(x,if(x%2,g = gcd(g,i)); x>>=1; i++); (g); }; \\ (Uses the program given in A052331) - Antti Karttunen, Feb 18 2023

Formula

For all n >= 1, a(A050376(n)) = n. - Antti Karttunen, Feb 18 2023

Extensions

Secondary definition added by Antti Karttunen, Feb 18 2023

A176525 Fermi-Dirac semiprimes: products of two distinct terms of A050376.

Original entry on oeis.org

6, 8, 10, 12, 14, 15, 18, 20, 21, 22, 26, 27, 28, 32, 33, 34, 35, 36, 38, 39, 44, 45, 46, 48, 50, 51, 52, 55, 57, 58, 62, 63, 64, 65, 68, 69, 74, 75, 76, 77, 80, 82, 85, 86, 87, 91, 92, 93, 94, 95, 98, 99, 100, 106, 111, 112, 115, 116, 117, 118, 119, 122
Offset: 1

Views

Author

Vladimir Shevelev, Apr 19 2010, Apr 20 2010

Keywords

Comments

The sequence essentially differs from A000379 beginning with a(108)=212 (not 210). All squarefree terms of A001358 are in the sequence.

References

  • Vladimir S. Shevelev, Multiplicative functions in the Fermi-Dirac arithmetic, Izvestia Vuzov of the North-Caucasus region, Nature sciences, Vol. 4 (1996), pp. 28-43 [Russian].

Crossrefs

Programs

  • Mathematica
    Select[Range[120], Plus @@ DigitCount[Last /@ FactorInteger[#], 2, 1] == 2 &] (* Amiram Eldar, Nov 27 2020 *)

Formula

If a(n)=u*v, uA050376 "Fermi-Dirac primes", then A064380(a(n))=a(n)-u-v+1+Sum{i>=1}(-1)^(i-1)*floor(v/u^i).

Extensions

Effectively duplicate content (due to duplicate referenced sequence) removed by Peter Munn, Dec 19 2019

A228520 a(n) is the smallest number such that if x >= a(n), then pi^*(x) - pi^*(x/2) >= n, where pi^*(x) is the number of terms of A050376 <= x.

Original entry on oeis.org

2, 3, 11, 16, 23, 41, 47, 59, 67, 71, 79, 101, 107, 109, 127, 149, 167, 169, 179, 181, 227, 229, 233, 239, 256, 263, 269, 281, 283, 307, 347, 349, 359, 367, 373, 401, 409, 419, 431, 433, 439, 461, 487, 491, 521, 569, 587, 593, 599, 601, 607, 617, 641, 643, 647
Offset: 1

Views

Author

Vladimir Shevelev, Aug 24 2013

Keywords

Comments

The sequence is a Fermi-Dirac analog of Ramanujan numbers (A104272), since terms of A050376 play a role of primes in Fermi-Dirac arithmetic (see comments in A050376).

Crossrefs

Cf. A104272.

Formula

a(n)<= R_n, where R_n is the n-th Ramanujan number (A104272); a(n)~A000040(2*n) as n goes to infinity.

Extensions

More terms from Peter J. C. Moses

A240906 Maximal A050376-factors in factorization of factorials over distinct terms of A050376.

Original entry on oeis.org

2, 3, 4, 5, 16, 81, 256, 65536, 4294967296, 18446744073709551616, 340282366920938463463374607431768211456, 115792089237316195423570985008687907853269984665640564039457584007913129639936
Offset: 2

Views

Author

Vladimir Shevelev, Apr 14 2014

Keywords

Comments

In contrast to prime power factorization of n! where maximal prime does not exceed n, in the considered factorization of n!, the maximal A050376-factor could be much greater than n. For example, for n=18, it is 65536.
Essentially, a(n) is A177334 with duplicates removed. - Ivan Neretin, May 31 2016

Examples

			81 is in the sequence, since 9! = 2*4*5*7*16*81.
		

Crossrefs

Extensions

a(9) from A177334 by Robert Price, Aug 31 2019
a(10)-a(12) from Amiram Eldar, Sep 17 2019

A241123 Smallest k such that the factorization of k! over distinct terms of A050376 contains exactly n primes.

Original entry on oeis.org

2, 3, 5, 13, 17, 21, 23, 37, 33, 42, 43, 56, 59, 57, 75, 84, 99, 101, 105, 109, 123, 119, 133, 139, 157, 162, 163, 182, 186, 183, 207, 208, 222, 219, 235, 220, 255, 257, 263, 268, 267, 303, 305, 307, 316, 315, 340, 344, 341, 343, 383, 385, 387, 397, 411, 425
Offset: 1

Views

Author

Vladimir Shevelev, Apr 16 2014

Keywords

Examples

			Factorization of 5! over distinct terms of A050376 is 5! = 2*3*4*5. Thus 5 is the smallest k such that such a factorization contains 3 primes: 2,3,5. So a(3)=5.
		

References

  • V. S. Shevelev, Multiplicative functions in the Fermi-Dirac arithmetic, Izvestia Vuzov of the North-Caucasus region, Nature sciences 4 (1996), 28-43 (Russian; MR 2000f: 11097, pp. 3912-3913).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Mod[e, 2]; b[1] = 0; b[n_] := Plus @@ (f @@@ FactorInteger[n]); m = 56; v = Table[0, {m}]; c = 0; p = 1; n = 2; While[c < m, p *= n; i = b[p]; If[i <= m && v[[i]] == 0, c++; v[[i]] = n]; n++]; v (* Amiram Eldar, Sep 17 2019 *)
  • PARI
    nbp(n) = {f = factor(n); sum (i=1, #f~, f[i,2] % 2);}
    a(n) = {k = 1; while(nbp(k!) != n, k++); k;} \\ Michel Marcus, Apr 27 2014

Extensions

More terms from Peter J. C. Moses, Apr 17 2014
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