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.

User: Tomohiro Yamada

Tomohiro Yamada's wiki page.

Tomohiro Yamada has authored 37 sequences. Here are the ten most recent ones:

A369205 Numbers m such that A188999(A034448(m)) = k*m for some k, where A034448 and A188999 are respectively the unitary and the bi-unitary sigma function.

Original entry on oeis.org

1, 2, 9, 10, 15, 18, 21, 30, 40, 42, 60, 120, 288, 567, 630, 720, 756, 1023, 1134, 1428, 2046, 2160, 2268, 2520, 3024, 3276, 3570, 4092, 6048, 8184, 8925, 9240, 11424, 11550, 15345, 17850, 18144, 30690, 35700, 46200, 57120, 85680, 147312, 285600, 491040, 556920
Offset: 1

Author

Tomohiro Yamada, Jan 16 2024

Keywords

Examples

			A034448(18) = 4 * 10 = 40 and A188999(40) = 15 * 6 = 90 = 5 * 18, so 18 is a term with k = 5.
		

Crossrefs

Cf. A038843 (analog for A034448(A034448(m))), A318175 (analog for A188999(A188999(m))).
Cf. A369204 (analog for A034448(A188999(m))).

Programs

  • PARI
    a034448(n) = {my(f,i,p,e);f=factor(n);for(i=1,#f~,p=f[i,1];e=f[i,2];f[i,1]=p^e+1;f[i,2]=1);factorback(f)};
    a188999(n) = {my(f,i,p,e);f=factor(n);for(i=1,#f~,p=f[i,1];e=f[i,2];f[i,1]=if(e%2,(p^(e+1)-1)/(p-1),(p^(e+1)-1)/(p-1)-p^(e/2));f[i,2]=1);factorback(f)};
    isok(n) = (a188999(a034448(n))%n) == 0;

A369204 Numbers m such that A034448(A188999(m)) = k*m for some k, where A034448 and A188999 are respectively the unitary and the bi-unitary sigma function.

Original entry on oeis.org

1, 2, 8, 9, 10, 18, 24, 27, 30, 54, 165, 238, 288, 512, 656, 660, 864, 952, 1536, 1968, 2464, 2880, 4608, 4680, 13824, 14448, 14976, 16728, 19008, 19992, 23040, 29376, 60928, 152064, 155520, 172368, 279552, 474936, 746928, 1070592, 1114560, 1524096, 1703520
Offset: 1

Author

Tomohiro Yamada, Jan 16 2024

Keywords

Examples

			A188999(18) = 4 * 10 = 40 and A034448(40) = 9 * 6 = 54 = 3 * 18, so 18 is a term with k = 3.
		

Crossrefs

Cf. A038843 (analog for A034448(A034448(m))), A318175 (analog for A188999(A188999(m))).
Cf. A369205 (analog for A188999(A034448(m))).

Programs

  • PARI
    a034448(n) = {my(f,i,p,e);f=factor(n);for(i=1,#f~,p=f[i,1];e=f[i,2];f[i,1]=p^e+1;f[i,2]=1);factorback(f)};
    a188999(n) = {my(f,i,p,e);f=factor(n);for(i=1,#f~,p=f[i,1];e=f[i,2];f[i,1]=if(e%2,(p^(e+1)-1)/(p-1),(p^(e+1)-1)/(p-1)-p^(e/2));f[i,2]=1);factorback(f)};
    isok(n) = (a034448(a188999(n))%n) == 0;

A368424 Numbers k such that gcd(A019320(k), A019321(k)) > 1.

Original entry on oeis.org

4, 11, 18, 20, 23, 28, 35, 43, 48, 52, 83, 95, 100, 119, 131, 138, 148, 155, 162, 166, 172, 179, 191, 196, 204, 210, 214, 239, 251, 253, 268, 292, 299, 300, 316, 323, 342, 359, 371, 378, 388, 419, 431, 443, 460, 463, 491, 500, 508, 515, 537, 556, 564, 575
Offset: 1

Author

Tomohiro Yamada, Dec 24 2023

Keywords

Comments

The corresponding greatest common divisors are given in A368425.

Examples

			a(1) = 4 since A019320(4) = 5 and A019321(4) = 10.
		

Crossrefs

Cf. A019320, A019321, A191609 (prime factors of such gcds), A368425.

Programs

  • Maple
    select(k -> igcd(numtheory:-cyclotomic(k,2),
    numtheory:-cyclotomic(k,3)) > 1, [$1..1000]); # Robert Israel, Dec 26 2023
  • Mathematica
    Select[Range[600],GCD[Cyclotomic[#,2],Cyclotomic[#,3]]>1&] (* Stefano Spezia, Dec 26 2023 *)
  • PARI
    for(n=1,1000,if(gcd(polcyclo(n,2),polcyclo(n,3))>1,print1(n,", ")))

A368425 The corresponding greatest common divisors to A368424(n).

Original entry on oeis.org

5, 23, 19, 5, 47, 29, 71, 431, 97, 53, 167, 191, 505, 239, 263, 139, 149, 311, 163, 499, 173, 359, 383, 197, 409, 211, 643, 479, 503, 23, 269, 293, 599, 1201, 317, 647, 19, 719, 743, 379, 389, 839, 863, 887, 461, 11113, 983, 5, 509, 1031, 4297, 557, 1129
Offset: 1

Author

Tomohiro Yamada, Dec 24 2023

Keywords

Examples

			a(2) = 23 since gcd(A019320(A368424(2)), A019321(A368424(2))) = gcd(2047, 88573) = 23.
		

Crossrefs

Cf. A019320, A019321, A191609 (primes dividing some term of this sequence), A368424.

Programs

  • Maple
    subs(1=NULL, [seq(igcd(numtheory:-cyclotomic(n,2), numtheory:-cyclotomic(n,3)),n=1..1000)]); # Robert Israel, Dec 26 2023
  • Mathematica
    Select[GCD[Cyclotomic[Range[600], 2], Cyclotomic[Range[600], 3]],#>1&] (* Stefano Spezia, Dec 26 2023 *)
  • PARI
    for(n=1,1000,m=gcd(polcyclo(n,2),polcyclo(n,3));if(m>1,print1(m,", ")))

A362531 The smallest integer m such that m mod 2k >= k for k = 1, 2, 3, ..., n.

Original entry on oeis.org

1, 3, 3, 15, 15, 47, 95, 95, 287, 335, 1199, 1199, 1295, 2015, 2879, 2879, 2879, 2879, 2879, 2879, 2879, 43199, 211679, 211679, 211679, 211679, 211679, 211679, 211679, 211679, 3084479, 3084479, 3084479, 3084479, 3084479, 3084479, 302702399, 469909439
Offset: 1

Author

Tomohiro Yamada, Apr 24 2023

Keywords

Comments

Take the square array A(k, l) with k= 1, 2, ... and l = 0, 1, ... such that for each k, A(k, l) takes k zeros and then k ones alternately:
0, 1, 0, 1, 0, 1, 0, 1, ...
0, 0, 1, 1, 0, 0, 1, 1, ...
0, 0, 0, 1, 1, 1, 0, 0, ...
...
Then the a(n)-th column is the first column which begins with n ones.

Examples

			a(3) = 3 since 3 mod 2 = 1, 3 mod 4 = 3 >= 2, 3 mod 6 = 3 (but 3 mod 8 = 3 < 4) while 1 mod 4 = 1 < 2, 2 mod 2 = 0 < 1.
		

Crossrefs

Cf. A053664.

Programs

  • PARI
    a(n)={my(m);m=1;while(vecmin(vector(n,j,(m%(2*j))/j))<1,m=m+1);m}
    
  • PARI
    n=1;for(m=1,10^9,while(vecmin(vector(n,j,(m%(2*j))/j))>=1,print(n," ",m);n=n+1))
    
  • Python
    def A362531(n):
        m = 1
        while True:
            for k in range(n,0,-1):
                if (l:=k-m%(k<<1))>0:
                    break
            else:
                return m
            m += l # Chai Wah Wu, Jun 21 2023

Extensions

a(37)-a(38) from Yifan Xie, Apr 24 2023

A362532 The smallest positive integer m such that m mod 2k < k for k = 1, 2, 3, ..., n.

Original entry on oeis.org

2, 4, 8, 8, 24, 24, 72, 144, 144, 144, 384, 384, 2160, 2160, 2160, 6720, 54240, 57600, 131040, 131040, 131040, 131040, 612000, 612000, 612000, 612000, 612000, 776160, 776160, 776160, 6219360, 23184000, 28627200, 28627200, 28627200, 28627200, 28627200
Offset: 1

Author

Tomohiro Yamada, Apr 24 2023

Keywords

Comments

Take the square array A(k, l) with k = 1, 2, ... and l = 0, 1, ... such that for each k, A(k, l) takes k zeros and then k ones alternately:
0, 1, 0, 1, 0, 1, 0, 1, ...
0, 0, 1, 1, 0, 0, 1, 1, ...
0, 0, 0, 1, 1, 1, 0, 0, ...
...
Then the a(n)-th column is the first column which begins with n zeros except the 0th column.

Examples

			a(2) = 4 since 4 mod 2 = 0, 4 mod 4 = 0 (but 4 mod 6 = 4 >= 3) while 1 mod 2 = 1, 2 mod 4 = 2, 3 mod 2 = 1.
		

Crossrefs

Cf. A053664.

Programs

  • PARI
    a(n)=my(m);m=1;while(vecmax(vector(n,j,(m%(2*j))/j))>=1,m=m+1);m
    
  • PARI
    n=1;for(m=1,10^9,while(vecmax(vector(n,j,(m%(2*j))/j))<1,print(n," ",m);n=n+1))
    
  • PARI
    a(n,startAt=1)=if(n<5, return(2^min(n,3))); my(s=if(n>146,70274254050, n>108,10039179150, n>94,436486050, n>65,22972950, n>50,417690, n>46,8190, n>27,630, n>17,90, n>12,30, 6)<=k, next(2))); return(m)) \\ Charles R Greathouse IV, Apr 28 2023
    
  • Python
    from itertools import count, islice
    def agen(): # generator of terms
        m = 1
        for n in count(1):
            while not all(m%(2*k) < k for k in range(1, n+1)): m += 1
            yield m
    print(list(islice(agen(), 37))) # Michael S. Branicky, Apr 24 2023

Formula

Trivial bounds: a(n-1) <= a(n) <= 2*A003418(n). - Charles R Greathouse IV, May 08 2023

A355298 Primes p such that q divides p + 1, r divides q^2 + q + 1, s divides r^2 + r + 1, and p divides s^2 + s + 1 for some primes q, r, and s.

Original entry on oeis.org

3, 13, 61, 127, 399403
Offset: 1

Author

Tomohiro Yamada, Jun 28 2022

Keywords

Comments

There are no other terms below 2^24.
If rad(n)^2 = sigma(n), where rad(n) = A007927(n) is the largest squarefree number dividing n and sigma(n) = A000203(n) is the sum of divisors of n, and there exists just one odd prime factor p dividing n exactly once, then p must belong to A354427 or this sequence.

Examples

			61 is a term since 61 + 1 = 2 * 31, 31^2 + 31 + 1 = 3 * 331, 3^2 + 3 + 1 = 13, and 13^2 + 13 + 1 = 3 * 61.
		

Crossrefs

Programs

  • PARI
    is(p)={my(W, V1, V2, V3, V4, q1, q2, q3, q4, i1, i2, i3, i4, l1, l2, l3, l4); W=0; V1=factor(p+1); l1=length(V1[, 1]); for(i1=1, l1, q1=V1[i1, 1]; V2=factor(q1^2+q1+1); l2=length(V2[, 1]); for(i2=1, l2, q2=V2[i2, 1]; V3=factor(q2^2+q2+1); l3=length(V3[, 1]); for(i3=1, l3, q3=V3[i3, 1]; V4=factor(q3^2+q3+1); l4=length(V4[, 1]); for(i4=1, l4, q4=V4[i4, 1];if(q4==p, W=[p, q1, q2, q3]))))); W}

A354428 Primes p such that q divides p^2 + p + 1, r divides q + 1 and p divides r^2 + r + 1 for some primes q and r.

Original entry on oeis.org

3, 7, 43, 73363, 1477111
Offset: 1

Author

Tomohiro Yamada, May 27 2022

Keywords

Comments

There are no other terms below 2^24.
The first five terms correspond to 7, 2, 79, 9829, and 5569 in A354426 respectively.
Similarly, these correspond to 13, 3 (or 19), 631, 1794067711, and 10855016833 in A354427 respectively.

Examples

			43 is a term since 43^2 + 43 + 1 = 3 * 631, 631 + 1 = 2^3 * 79, and 79^2 + 79 + 1 = 3 * 7^2 * 43.
		

Crossrefs

Cf. A354426 (r corresponding to primes p in this sequence), A354427 (q corresponding to primes p in this sequence).

Programs

  • PARI
    is(p)={my(W, V1, V2, V3, q1, q2, q3, i1, i2, i3, l1, l2, l3); W=0; V1=factor(p^2+p+1); l1=length(V1[, 1]); for(i1=1, l1, q1=V1[i1, 1]; V2=factor(q1+1); l2=length(V2[, 1]); for(i2=1, l2, q2=V2[i2, 1]; V3=factor(q2^2+q2+1); l3=length(V3[, 1]); for(i3=1, l3, q3=V3[i3, 1]; if(q3==p, W=[p, q1, q2])))); W}

A354427 Primes p such that q divides p + 1, r divides q^2 + q + 1 and p divides r^2 + r + 1 for some primes q and r.

Original entry on oeis.org

3, 13, 19, 631
Offset: 1

Author

Tomohiro Yamada, May 27 2022

Keywords

Comments

There are no other terms below 2^24.
1794067711 and 10855016833 are also terms, which arise from q = 9829 and q = 5569 in A354426 respectively, since
1794067711 + 1 = 2^8 * 23 * 31 * 9829, 9829^2 + 9829 + 1 = 3 * 439 * 73363, 73363^2 + 73363 + 1 = 3 * 1794067711 and
10855016833 + 1 = 2 * 17 * 5569 * 57329, 5569^2 + 5569 + 1 = 3 * 7 * 1477111, 1477111^2 + 1477111 + 1 = 3 * 67 * 10855016833.
If rad(n)^2 = sigma(n), where rad(n) = A007927(n) is the largest squarefree number dividing n and sigma(n) = A000203(n) is the sum of divisors of n, and there exists just one odd prime factor p dividing n exactly once, then p must belong to this sequence, or q divides p + 1, r divides q^2 + q + 1, s divides r^2 + r + 1, and p divides s^2 + s + 1 for some primes q, r, and s (Yamada, 2021). [Corrected by Tomohiro Yamada, Jun 05 2022]

Examples

			631 is a term since 631 + 1 = 2^3 * 79, 79^2 + 79 + 1 = 3 * 7^2 * 43, and 43^2 + 43 + 1 = 3 * 631.
		

Crossrefs

Cf. A354426 (q corresponding to primes p in this sequence), A354428 (r corresponding to primes p in this sequence).

Programs

  • PARI
    is(p)={my(W, V1, V2, V3, q1, q2, q3, i1, i2, i3, l1, l2, l3); W=0; V1=factor(p+1); l1=length(V1[, 1]); for(i1=1, l1, q1=V1[i1, 1]; V2=factor(q1^2+q1+1); l2=length(V2[, 1]); for(i2=1, l2, q2=V2[i2, 1]; V3=factor(q2^2+q2+1); l3=length(V3[, 1]); for(i3=1, l3, q3=V3[i3, 1]; if(q3==p, W=[p, q1, q2])))); W}

A354426 Primes p such that q divides p^2 + p + 1, r divides q^2 + q + 1 and p divides r + 1 for some primes q and r.

Original entry on oeis.org

2, 7, 79, 5569, 9829
Offset: 1

Author

Tomohiro Yamada, May 27 2022

Keywords

Comments

There are no other terms below 2^24.
There are no other terms below 10^8. - Lucas A. Brown, Aug 11 2024

Examples

			7 is a term since 7^2 + 7 + 1 = 3 * 19, 3^2 + 3 + 1 = 13 and 13 + 1 = 2 * 7.
		

Crossrefs

Programs

  • PARI
    is(p)={my(W, V1, V2, V3, q1, q2, q3, i1, i2, i3, l1, l2, l3); W=0; V1=factor(p^2+p+1); l1=length(V1[, 1]); for(i1=1, l1, q1=V1[i1, 1]; V2=factor(q1+1); l2=length(V2[, 1]); for(i2=1, l2, q2=V2[i2, 1]; V3=factor(q2^2+q2+1); l3=length(V3[, 1]); for(i3=1, l3, q3=V3[i3, 1]; if(q3==p, W=[p, q1, q2])))); W}