A060355 -id:A060355 - cp's OEIS frontend

A240591 The smaller of a pair of successive powerful numbers (A001694) without any prime number between them.

8, 25, 32, 121, 288, 675, 1331, 1369, 1936, 2187, 2700, 3125, 5324, 6724, 9800, 10800, 12167, 15125, 32761, 39200, 48668, 70225, 79507, 88200, 97336, 107648, 143641, 156800, 212521, 228484, 235224, 280900, 312481, 332928, 456968, 465124, 574564, 674028, 744769, 829921, 830297, 857476, 877952, 940896

Offset: 1

Antonio Roldán, Apr 08 2014

nonn

			25 is in the sequence because A001694(6)=25, A001694(7)=27, without primes between them.

Amiram Eldar, Table of n, a(n) for n = 1..373 (terms below 10^15; terms 1..103 from Amiram Eldar, terms 104..235 from Chai Wah Wu)

Supersequence of A060355.

Cf. A001694, A240590.

Select[Partition[Join[{1},Select[Range[10^6],Min@FactorInteger[#][[All, 2]]> 1&]],2,1],PrimePi[#[[1]]]==PrimePi[#[[2]]]&][[All,1]] (* Harvey P. Dale, Mar 28 2018 *)

ispowerful(n)={local(h);if(n==1,h=1,h=(vecmin(factor(n)[, 2])>1));return(h)}
nextpowerful(n)={local(k);k=n+1;while(!ispowerful(k),k+=1);return(k)}
{for(i=1,10^6,if(ispowerful(i),if(nextprime(i)>=nextpowerful(i),print1(i, ", "))))}

from itertools import count, islice
from math import isqrt
from sympy import mobius, integer_nthroot, nextprime
def A240591_gen(): # generator of terms
    def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
    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
    def f(x):
        c, l, j = x-squarefreepi(integer_nthroot(x,3)[0]), 0, isqrt(x)
        while j>1:
            k2 = integer_nthroot(x//j**2,3)[0]+1
            w = squarefreepi(k2-1)
            c -= j*(w-l)
            l, j = w, isqrt(x//k2**3)
        return c+l
    m = 1
    for n in count(2):
        k = bisection(lambda x:f(x)+n,m,m)
        if nextprime(m) > k:
            yield m
        m = k
A240591_list = list(islice(A240591_gen(),30)) # Chai Wah Wu, Sep 14 2024

A359747 Numbers k such that k*(k+1) has in its canonical prime factorization mutually distinct exponents.

Original entry on oeis.org

1, 3, 4, 7, 8, 16, 24, 27, 31, 48, 63, 71, 72, 107, 108, 124, 127, 199, 242, 243, 256, 400, 431, 432, 499, 512, 576, 647, 783, 863, 967, 971, 1024, 1151, 1152, 1372, 1567, 1600, 1999, 2187, 2311, 2400, 2591, 2592, 2887, 2916, 3087, 3136, 3456, 3887, 3888, 3968, 4000

Offset: 1

Amiram Eldar, Jan 13 2023

nonn

Equivalently, numbers k such that A002378(k) = k*(k+1) is a term of A130091.
Equivalently, numbers k such that the multisets of exponents in the prime factorizations of k and k+1 are disjoint and each have distinct elements.
Either k or k+1 is a powerful number (A001694). Except for k=8, are there terms k such that both k and k+1 are powerful (i.e., terms that are also in A060355)? None of the terms A060355(n) for n = 2..39 is in this sequence.

			3 is a term since 3*4 = 12 = 2^2 * 3^1 has 2 distinct exponents in its prime factorization: 1 and 3.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Subsequence of A130091 and A342028.
A359748 is a subsequence.
Cf. A001694, A002378, A060355.

q[n_] := UnsameQ @@ (FactorInteger[n*(n+1)][[;; , 2]]); Select[Range[4000], q]

is(n) = { my(e = factor(n*(n+1))[, 2]); #Set(e) == #e; }

A227297 Suppose that (m, m+1) is a pair of consecutive powerful numbers as defined by A001694. This sequence gives the values of m for which neither m nor m+1 are perfect squares.

Original entry on oeis.org

12167, 5425069447, 11968683934831, 28821995554247, 48689748233307, 161461422688535037152, 3887785221910670811499

Offset: 1

Ant King, Jul 07 2013

a(1) to a(5) were found by Jaroslaw Wroblewski, who also proved that this sequence is infinite (see link to Problem 53 below). However, there are no more terms less than 500^6 = 1.5625*10^16.

A subsequence of A060355 and of A001694.

			12167 is a term because (12167, 12168) are a pair of consecutive powerful numbers, neither of which are perfect squares.
235224 is not a term because although (235224, 235225) are a pair of consecutive powerful numbers, the larger member of the pair is a square number (= 485^2).

Richard K. Guy, Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, (1994), pp. 70-74. (See Powerful numbers, section B16.)

Solomon W. Golomb, Powerful numbers, Amer. Math. Monthly, Vol. 77, No. 8 (October 1970), 848-852.
Carlos Rivera, Problem 53: Powerful numbers revisited, The Prime Puzzles & Problems Connection.
David T. Walker, Consecutive integer pairs of powerful numbers and related Diophantine equations, Fibonacci Quart., Vol. 14, No. 2 (1976), pp. 111-116.
Wikipedia, Powerful number.
Index entries for sequences related to powerful numbers.

Cf. A060355, A001694.

a(6)-a(7) from the b-file at A060355 added by Amiram Eldar, Mar 22 2025

A348122 Numbers k such that k and k+1 both have more nonunitary than unitary prime divisors (A348121).

Original entry on oeis.org

8, 288, 360, 675, 1224, 1331, 1368, 2196, 2400, 2600, 2808, 3024, 5328, 6075, 6859, 9408, 9800, 10647, 11448, 12167, 16128, 17199, 19844, 20448, 21024, 23275, 25920, 26568, 26900, 28899, 29791, 33524, 38024, 38808, 39600, 40400, 41624, 42875, 45324, 46224, 46475

Offset: 1

Amiram Eldar, Oct 01 2021

nonn

			8 is a term since 8 = 2^3 has one nonunitary prime divisor, 2, and no unitary prime divisors, and 8 + 1 = 9 = 3^2 has one nonunitary prime divisor, 3, and no unitary prime divisors.

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

Subsequence of A348121.
A060355 is a subsequence.
Similar sequence: A348119.

q[n_] := 2*Count[(e = FactorInteger[n][[;; , 2]]), 1] < Length[e]; Select[Range[5*10^5], q[#] && q[# + 1] &]

A355433 Numbers k such that k is sqrt(k)-smooth and k+1 is sqrt(k+1)-smooth.

Original entry on oeis.org

8, 24, 48, 49, 63, 80, 120, 125, 168, 175, 195, 224, 242, 288, 324, 350, 351, 360, 363, 374, 384, 399, 440, 441, 455, 475, 494, 512, 528, 539, 560, 575, 594, 624, 675, 714, 728, 735, 759, 832, 840, 874, 896, 935, 960, 968, 1000, 1014, 1023, 1044, 1053, 1088, 1104

Offset: 1

Amiram Eldar, Jul 02 2022

nonn

Numbers k such that k and k+1 are both in A048098.

This sequence is infinite: if p is an odd prime then p^2-1 is a term.

			8 is a term since 8 is sqrt(8)-smooth (2^2 <= 8) and 9 is sqrt(9)-smooth (3^2 <= 9).

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

Cf. A048098, A355434.

Subsequences: A084920 \ {3}, A060355, A348119.

smQ[n_] := FactorInteger[n][[-1, 1]]^2 <= n; Select[Range[1000], smQ[#] && smQ[# + 1] &]

A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

Original entry on oeis.org

0, 682, 1268860318, 1459639851109444, 2360712083917682, 86149711981264908618, 4392100110703410665318, 8171493471761113423918890682, 15203047261220215902863544865414318, 5484296027914919579181500526692857773246, 28285239023397517753374058381589688919682, 12439333951782387734360136352377558500557329868

Offset: 1

Michel Lagneau, Feb 27 2010

nonn

This sequence is infinite. The fundamental solution of m^2 + 1 = x^2 y^3 is (m,x,y) = (682,61,5), which means the Pellian equation m^2 - 125x^2 = -1 has the solution (m,x) = (682,61) = (m(1),x(1)). This Pellian equation admits an infinity of solutions (m(2k+1),x(2k+1)), k=1,2,..., given by the following recursive relation, starting with m(1)=682, x(1)= 61: m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1).
Squares of these terms are in A060355, since both a(n)^2 and a(n)^2 + 1 are powerful (A001694). - Charles R Greathouse IV, Nov 16 2012
It appears that y = A077426. - Robert G. Wilson v, Nov 16 2012
Also m^2 + 1 is powerful. Other solutions arise from solutions x to x^2 - k^3*y^2 = -1. - Georgi Guninski, Nov 17 2012
Although it is believed that the b-file is complete for all terms m < 10^100, the search only looked for y < 100000. - Robert G. Wilson v, Nov 17 2012

			For m=682, m^2 + 1 = 465125 = 61^2 * 5^3.

Albert H. Beiler, "The Pellian" (Chap. 22), Recreations in the Theory of Numbers, 2nd ed. NY: Dover, 1966.
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 108.

Robert G. Wilson v, Table of n, a(n) for n = 1..292. [This list has not been proved to be complete! - _N. J. A. Sloane_, Nov 17 2012]
E. E. Whitford, The Pell equation, New York, 1912.
Wikipedia, Pell's equation

Pellian equation: A006704, A006702, A006705, A006703, A003153, A001571, A001570.

C:=array(0..20,0..20):C[1,1]=1: C[2,1]=1: n1:=682:x1:=61:for nn from 1 by 2 to 15 do:s:=0:for i from 2 to 15 do:for j from 1 to i do:C[i,j]:= C[i-1,j] + C[i-1,j-1]: od:od:for n from 1 by 2 to nn+1 do:s:=s + C[nn+1,n] * n1^(nn-n+1)*x1^(n-1)*125^((n-1)/2):od:print (s):od: # Michel Lagneau
# 2nd program R. J. Mathar, Mar 16 2016:
# print (nonsorted!) all solutions of A175155 up to search limit
with(numtheory):
# upper limit for solutions n
nsearchlim := 10^40 :
A175155y := proc(y::integer)
    local disc;
    disc := y^3 ;
    cfrac(sqrt(disc),periodic,quotients) ;
end proc:
for y from 2 do
    if issqrfree(y) then
        # find continued fraction for x^2-(y^3=disc)*y^2=-1, sqrt(disc)
        cf := A175155y(y) ;
        nlen :=  nops(op(2,cf)) ;
        if type(nlen,odd) then
            # fundamental solution
            fuso := numtheory[nthconver](cf,nlen-1) ;
            fusolx := numer(fuso) ;
            fusoly := denom(fuso) ;
            solx := fusolx ;
            soly := fusoly ;
            while solx <= nsearchlim do
                rhhs := solx^2-y^3*soly^2 ;
                if rhhs = -1 then
                    # print("n=",solx,"x=",soly,"y=",y^3) ;
                    print(solx) ;
                end if;
                # solutions from fundamental solution
                tempx := fusolx*solx+y^3*fusoly*soly ;
                tempy := fusolx*soly+fusoly*solx ;
                solx := tempx ;
                soly := tempy ;
            end do;
        end if;
    fi;
end do:

nmax = 10^50; ymax = 100; instances = 10; fi[y_] := n /. FindInstance[0 <= n <= nmax && x > 0 && n^2 + 1 == x^2*y^3, {n, x}, Integers, instances]; yy = Select[Range[1, ymax, 2], !IntegerQ[Sqrt[#]] && OddQ[ Length[ ContinuedFraction[Sqrt[#]][[2]]]]&]; Join[{0}, fi /@ yy // Flatten // Union // Most] (* Jean-François Alcover, Jul 12 2017 *)

is(n)=ispowerful(n^2+1) \\ Charles R Greathouse IV, Nov 16 2012

m(1)=682, x(1) = 61 and m(2k+1) + x(2k+1)*sqrt(125) = (m(1) + x(1)*sqrt(125))^(2k+1) m(2k+1) = C(2k+1,0) * m(1)^(2k+1) + C(2k+1,2)*m(1)^(2k-1)*x(1)^2 + ...

Added condition that x and y must be positive. Added missing initial term 0. Added warning that b-file has not been proved to be correct - there could be missing entries. - N. J. A. Sloane, Nov 17 2012

A178811 The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).

Original entry on oeis.org

1, 2, 4, 33, 8, 10093613546512321, 16, 28375, 1309, 32, 36, 7939375, 932537185321, 64

Offset: 1

Will Nicholes, Jun 16 2010

From Bernard Schott, Feb 17 2021: (Start)
The corresponding lengths of these longest runs of consecutive integers are in A178810.
If a(n) = 2^k for some k <> 1, then a(n) = A025487(n) and A178810(n) = 1; for k = 1, a(2) = A025487(2) = A178810(2) = 2 because there exists a run of two consecutive primes (2,3).
a(18) = 128, a(22) = 203433. [corrected by Jon E. Schoenfield, Nov 30 2023] (End)
From Jon E. Schoenfield, Dec 02 2023: (Start)
a(16) = 3302209375 = 5^5 * 1056707. (3302209376 = 2^5 * 103194043, 3302209377 = 3^5 * 13589339.) No run of four consecutive integers of the form p^5 * q with p,q distinct primes can exist: one of the two even numbers would be 32*q and the other would be 2*p^5, and they would differ by 2, yielding either (1) 2*p^5 + 2 = 32*q -> p^5 + 1 = 16*q -> (p^4 - p^3 + p^2 - p + 1)*(p+1) = 16*q, so p+1 = 16, but then p = 15 would not be a prime, or (2) 2*p^5 - 2 = 32*q -> p^5 - 1 = 16*q -> (p^4 + p^3 + p^2 + p + 1)*(p-1) = 16*q, so p-1 = 16, so p = 17, but then the odd number between 2*p^5 and 32*q would be 2*17^5 - 1 = 2839713 = 3 * 37 * 25583 (which would not have the required prime signature).
a(17) <= 921198089181020748838245 (which starts a run of seven consecutive integers of the form p^3*q*r; no run of eight or more can exist, as any set of eight consecutive integers includes an odd multiple of 4).
a(n) = A025487(n) if A025487(n) is a proper power (i.e., a number of the form b^e where b,e > 1). (Thus a(3) = 4, a(5) = 8, a(7) = 16, a(10) = 32, a(11) = 36, a(14) = 64, a(18) = 128, a(19) = 144, a(23) = 216, a(25) = 256, a(32) = 512, a(33) = 576, a(38) = 900, a(40) = 1024, a(44) = 1296, a(48) = 1728, a(51) = 2048, a(53) = 2304.)
Conjecture: a(n) = A025487(n) if A025487(n) is a powerful number (A001694); i.e., if A025487(n) is a powerful number, then there exists no run of two or more consecutive integers with the same prime signature as that of A025487(n). (E.g., if this conjecture holds, a(15) = 72 (cf. A367781), a(26) = 288, a(30) = 432, a(37) = 864, a(42) = 1152, a(49) = 1800.) (End)

			For n = 3, A025487(3) = 4, corresponding to a prime signature of {2}. Since the maximum number of consecutive integers with that prime signature is 1, a(3) is 4, the smallest integer that starts a "run" of 1.
A025487(4) = 6 whose prime signature is {1,1}; a(4) = 33 because 33 is the smallest integer where starts a run of A178810(4) = 3 consecutive integers with prime signature {1,1}: (33=3*11, 34=2*17, 35=5*7). - _Bernard Schott_, Feb 16 2021

Diophante, A1845, Les squelettes (in French).

Cf. A001694, A025487, A060355, A178810 (maximum size of such runs), A141621.

Minor edits by Ray Chandler, Jul 29 2010
a(6) corrected by Bobby Jacobs, Sep 25 2016
a(12) from Hugo van der Sanden, May 20 2019
a(13)-a(14) from Bernard Schott, Feb 16 2021

A190840 a(n+1) = 4a(n)(a(n)+1) for a(0) = 1.

Original entry on oeis.org

1, 8, 288, 332928, 443365544448, 786292024016459316676608, 2473020588127600939387543243786675530709484249088

Offset: 0

Alexander Zhukov, Aug 08 2011

nonn

For n>0, subsequence of A132592: both a(n)/2 and a(n)+1 are squares.
All terms (n > 0) are divisible by 8, yielding all terms of A185097, which is indexed from n=1, thus having the first term A185097(1) = 1.
The next term has 98 digits. - Harvey P. Dale, Jan 01 2014
For n>0, subsequence of A060355: both a(n) and a(n)+1 are powerful numbers. - Bernard Schott, Apr 24 2023

Cf. A000217, A001601, A060355, A132592, A185097.

NestList[4#(#+1)&,1,7] (* Harvey P. Dale, Jan 01 2014 *)

a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.
a(n) = sinh(2^(n-2)*arccosh(17))^2. - Alexander R. Povolotsky, Aug 14 2011
a(n) = 8*A185097(n) for n > 0. - Alexander R. Povolotsky, Aug 14 2011
a(n) = (1 + sqrt(2))^(2^(n+1))/4 + (1 - sqrt(2))^(2^(n+1))/4 - 1/2. Therefore 2*a(n) + 1 = A001601(n+1). - Bruno Berselli, Feb 01 2017

A355462 Powerful numbers divisible by exactly 2 distinct primes.

Original entry on oeis.org

36, 72, 100, 108, 144, 196, 200, 216, 225, 288, 324, 392, 400, 432, 441, 484, 500, 576, 648, 675, 676, 784, 800, 864, 968, 972, 1000, 1089, 1125, 1152, 1156, 1225, 1296, 1323, 1352, 1372, 1444, 1521, 1568, 1600, 1728, 1936, 1944, 2000, 2025, 2116, 2304, 2312, 2500

Offset: 1

Amiram Eldar, Jul 03 2022

nonn

First differs from A286708 at n = 25.
Number of the form p^i * q^j, where p != q are primes and i,j > 1.
Numbers k such that A001221(k) = 2 and A051904(k) >= 2.
The possible values of the number of the divisors (A000005) of terms in this sequence is any composite number that is not 8 or twice a prime (A264828 \ {1, 8}).
675 = 3^3*5^2 and 676 = 2^2*13^2 are 2 consecutive integers in this sequence. There are no other such pairs below 10^22 (the lesser members of such pairs are terms of A060355).

			36 is a term since 36 = 2^2 * 3^2.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index to sequences related to prime signature.

Intersection of A001694 and A007774.
Subsequence of A286708.
Subsequences: A085986, A143610, A162142, A179646, A179666, A179671, A179689, A179694, A179699, A179702, A179705, A189988, A189990, A189991, A190464, A190465, A303661.
Cf. A000005, A001221, A051904, A060355, A136141, A264828.

Select[Range[2500], Length[(e = FactorInteger[#][[;; , 2]])] == 2 && Min[e] > 1 &]

is(n) = {my(f=factor(n)); #f~ == 2 && vecmin(f[,2]) > 1};

Sum_{n>=1} 1/a(n) = ((Sum_{p prime} (1/(p*(p-1))))^2 - Sum_{p prime} (1/(p^2*(p-1)^2)))/2 = 0.1583860791... .

A359749 Numbers k such that k and k+1 do not share a common exponent in their prime factorizations.

Original entry on oeis.org

1, 3, 4, 7, 8, 9, 15, 16, 24, 25, 26, 27, 31, 32, 35, 36, 48, 63, 64, 71, 72, 81, 100, 107, 108, 120, 121, 124, 125, 127, 128, 143, 144, 168, 169, 195, 196, 199, 200, 215, 216, 224, 225, 242, 243, 255, 256, 287, 289, 323, 342, 361, 391, 392, 399, 400, 431, 432, 440

Offset: 1

Amiram Eldar, Jan 13 2023

nonn

Either k or k+1 is a powerful number (A001694). Except for k=8, are there terms k such that both k and k+1 are powerful (i.e., terms that are also in A060355)? None of the terms A060355(n) for n = 2..39 is in this sequence.

A002496(k)-1, A078324(k)-1, A078325(k)-1, and A049533(k)^2 are terms for all k >= 1.

			3 is a term since 3 has the exponent 1 in its prime factorization, and 3 + 1 = 4 = 2^2 has a different exponent in its prime factorization, 2.

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

Subsequences: A000079, A000225 \ {0}, A075408, A359747.

Cf. A001694, A002496, A049533, A060355, A078324, A078325.

q[n_] := UnsameQ @@ Join @@ (Union[FactorInteger[#][[;; , 2]]]& /@ (n + {0, 1})); Join[{1}, Select[Range[400], q]]

lista(nmax) = {my(e1 = [], e2); for(n = 2, nmax, e2 = Set(factor(n)[,2]); if(setintersect(e1, e2) == [], print1(n-1, ", ")); e1 = e2); }

cp's OEIS Frontend

A240591 The smaller of a pair of successive powerful numbers (A001694) without any prime number between them.

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A359747 Numbers k such that k*(k+1) has in its canonical prime factorization mutually distinct exponents.

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A227297 Suppose that (m, m+1) is a pair of consecutive powerful numbers as defined by A001694. This sequence gives the values of m for which neither m nor m+1 are perfect squares.

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A348122 Numbers k such that k and k+1 both have more nonunitary than unitary prime divisors (A348121).

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A355433 Numbers k such that k is sqrt(k)-smooth and k+1 is sqrt(k+1)-smooth.

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A175155 Numbers m satisfying m^2 + 1 = x^2 * y^3 for positive integers x and y.

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A178811 The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).

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A190840 a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.

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A190840 a(n+1) = 4a(n)(a(n)+1) for a(0) = 1.