Richard Peterson has authored 10 sequences.
A377266
Primes p with the property that there exist nonnegative integers m,n such that m!*n! is congruent to either +1 or -1 mod p^2, with m + n = p - 1.
Original entry on oeis.org
2, 3, 5, 7, 11, 13, 17, 31, 47, 53, 59, 61, 71, 107, 137, 149, 173, 227, 251, 277, 313, 347, 349, 359, 367, 373, 409, 419, 443, 463, 467, 479, 491, 499, 521, 523, 541, 563, 577, 599, 607, 613, 617, 631, 643, 647, 677, 683, 739, 751, 757, 809, 811, 821, 823, 827, 829
Offset: 1
0!*4! + 1 = 5^2 and 4+0 = 5-1, so 5 is in the sequence.
1!*9! - 1 = 11^2*2999 and 1+9 = 10-1, so 11 is in the sequence.
0!*12! + 1 = 13^2*2834329 and 0+12 = 13-1, so 13 is in the sequence.
10!*6! + 1 = 17^2*83*108923 and 10+6=17-1, so 17 is in the sequence.
19!*39!-1 is divisible by 59^2 and 19+39=59-1, so 59 is in the sequence.
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filter:= proc(p) local m,n,A,v;
A:= Array(0..p);
A[0]:= 1:
for n from 1 to p do A[n]:= n*A[n-1] mod p^2 od:
for m from 0 to (p-1)/2 do
v:= A[m] * A[p-1-m] mod p^2;
if v = 1 or v = p^2-1 then return true fi;
od;
false
end proc:
select(filter, [seq(ithprime(i),i=1..150)]); # Robert Israel, Dec 30 2024
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isok(p)={if(isprime(p), for(m=0, p\2, my(t=(m!*(p-1-m)!%p^2)); if(t==1||t==p^2-1, return(1)))); 0} \\ Andrew Howroyd, Oct 22 2024
A362829
Positions in lexicographic order of odd partitions of sufficiently large numbers.
Original entry on oeis.org
1, 3, 7, 10, 15, 20, 27, 30, 39, 41, 51, 56, 69, 72, 75, 93, 95, 101, 123, 128, 132, 134, 160, 163, 166, 172, 176, 212, 214, 220, 227, 229, 273, 278, 282, 284, 291, 297, 353, 356, 359, 365, 369, 379, 382, 384, 453, 455, 461, 468, 470, 481, 483, 490, 579, 584
Offset: 1
a(1)=1 because 1+1+...+1 (k times) is the first partition in lexicographic order of any positive integer k, and it is odd.
a(2)=3 because 1+1+...+1(k-3 times)+3=k is the third partition of k lexicographically and it is odd.
A350813
a(n) is the least positive number k such that the product of the first n primes that are congruent to 1 (mod 4) is equal to y^2 - k^2 for some integer y.
Original entry on oeis.org
2, 4, 24, 38, 16, 588, 5782, 5528, 80872, 319296, 3217476, 32301914, 20085008, 166518276, 2049477188, 17443412442, 27905362944, 233647747282, 886295348972, 134684992249108, 98002282636962, 392994156083892, 5283713761100536, 76642755213473624, 923250078609721236
Offset: 1
For n=3, m = 5*13*17. The "middle" most nearly equal divisor and codivisor of m are y-k=17 and y+k=65, whence a(n) = (65 - 17)/2 = 24.
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from math import prod, isqrt
from itertools import islice
from sympy import sieve, divisors
def A350813(n):
m = prod(islice(filter(lambda p: p % 4 == 1, sieve),n))
a = isqrt(m)
d = max(filter(lambda d: d <= a, divisors(m,generator=True)))
return (m//d-d)//2 # Chai Wah Wu, Mar 29 2022
Terms corrected by and more terms from
Jinyuan Wang, Mar 17 2022
A348635
a(n) is the smallest positive number k coprime to (2n+1)!! such that (2n+1)!! + k^2 is a square.
Original entry on oeis.org
1, 1, 4, 4, 29, 17, 436, 356, 569, 1847, 27704, 72944, 1283333, 726079, 23833532, 45232276, 302068799, 616565857, 26369361188, 23157514888, 70991664061, 505527042479, 1150735735948, 13238389944712, 58668785675111, 209280259070287, 7809609503808088, 530566746979816
Offset: 1
a(5)=29 since 106^2 - 29^2 = 10395 = 3*5*7*9*11 and 29 is relatively prime to 10395 and is as small as possible.
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df(n) = (2*n)! / n! / 2^n; \\ A001147
a(n) = my(d=df(n+1), k=1); while (!((gcd(d,k)==1) && issquare(d+k^2)), k++); k; \\ Michel Marcus, Jan 06 2022
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df(n) = (2*n)! / n! / 2^n; \\ A001147
a(n) = my(d=df(n+1), m=sqrtint(d), k); while (!(issquare(m^2-d, &k) && gcd(d,k)==1), m++); k; \\ Michel Marcus, Jan 06 2022
A349708
a(n) is the smallest positive number k such that (product of the first n odd primes) + k^2 is a square.
Original entry on oeis.org
1, 1, 4, 1, 19, 53, 58, 97, 181, 4244, 2122, 31126, 16451, 297392, 2444006, 622249, 2909047, 216182072, 62801719, 769709491, 32522441312, 37859955467, 129549407177, 286721160343, 101419856449, 107709289064864, 72441253480727, 56099073382147, 5249126879235893
Offset: 1
a(4)=1 because the product of the first 4 odd primes, 3*5*7*11 = 1155, is 34^2 - 1. a(5)=19 because 15015=3*5*7*11*13=124^2-19^2, and no positive integer less than 19 will work in this situation.
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a(n) = my(k=1, p=prod(k=2, n+1, prime(k))); while (!issquare(k^2+p), k++); k; \\ Michel Marcus, Jan 10 2022
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from math import isqrt
from sympy import primorial, divisors
def A349708(n):
m = primorial(n+1)//2
a = isqrt(m)
d = max(filter(lambda d: d <= a, divisors(m,generator=True)))
return (m//d-d)//2 # Chai Wah Wu, Mar 29 2022
a(15)-a(26) and corrections to a(9) and a(11) from
Jinyuan Wang, Jan 07 2022
A341678
Irregular triangle read by rows: row n consists of all numbers x such that x^2 + y^2 = A006278(n), with 0 < x < y.
Original entry on oeis.org
1, 1, 4, 4, 9, 12, 23, 2, 19, 46, 67, 74, 86, 109, 122, 64, 103, 167, 191, 236, 281, 292, 359, 449, 512, 568, 601, 607, 664, 673, 743, 59, 132, 531, 581, 627, 876, 1008, 1284, 1588, 1659, 1723, 2092, 2136, 2317, 2373, 2736, 2757, 2803, 3072, 3164, 3333, 3469, 3704, 3821, 4028, 4077, 4136, 4371, 4596, 4668, 4712, 4851
Offset: 1
Triangle starts:
1,
1, 4,
4, 9, 12, 23,
2, 19, 46, 67, 74, 86, 109, 122,
64, 103, 167, 191, 236, 281, 292, 359, 449, 512, 568, 601, 607, 664, 673, 743,
...
In the second row, calculations are as follows. 5*13 is the product of the first two primes congruent to 1 (mod 4), and 65 = 1^2 + 8^2 = 4^2 + 7^2, so the second row is 1, 4.
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row(n) = {my(t=1, q=3, v=vector(2^n/2)); for(k=1, n, until(q%4==1, q=nextprime(q+1)); t*=q); q=0; for(k=1, #v, until(issquare(t-q^2), q++); v[k]=q); v; } \\ Jinyuan Wang, Mar 03 2021
A338085
a(n) is the cardinality of S(n), the subset of partitions of n such that there are enough smaller parts to add together to be greater than a larger part.
Original entry on oeis.org
0, 0, 0, 0, 1, 1, 5, 5, 12, 18, 30, 36, 65, 83, 120, 159, 225, 284, 395, 495, 665, 848, 1094, 1348, 1757, 2184, 2746, 3399, 4250, 5199, 6469, 7867, 9667, 11756, 14310, 17266, 20988, 25216, 30372, 36371, 43648, 52041, 62187, 73866, 87837, 104105, 123279, 145453
Offset: 1
A337674
Numbers k whose prime divisors are all less than or equal to the number of divisors of k.
Original entry on oeis.org
1, 2, 4, 6, 8, 9, 12, 16, 18, 20, 24, 27, 30, 32, 36, 40, 42, 45, 48, 50, 54, 56, 60, 64, 70, 72, 75, 80, 81, 84, 90, 96, 100, 105, 108, 112, 120, 126, 128, 132, 135, 140, 144, 150, 160, 162, 168, 180, 189, 192, 196, 198, 200, 210, 216, 220, 224, 225, 240, 243, 250
Offset: 1
42=a(17) is a term, since 2,3 and 7 are the prime divisors of 42, which has 8 divisors. 156=2^2*3*13 is not a term, since 13 is greater than 12, the number of divisors of 156.
A199768 has "strictly less", while this sequence has "less than or equal to".
A146982 does not include terms 42, 56, 132, 198, 220, 264, 308, 312, 330, ...
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Select[Range[250], FactorInteger[#][[-1, 1]] <= DivisorSigma[0, #] &] (* Amiram Eldar, Sep 22 2020 *)
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isok(m) = #select(x->(x>numdiv(m)), factor(m)[,1]) == 0; \\ Michel Marcus, Sep 22 2020
A333986
Greatest prime factor of A007497(n).
Original entry on oeis.org
2, 3, 2, 7, 2, 5, 3, 5, 7, 5, 7, 5, 127, 7, 8191, 5, 127, 13, 31, 127, 127, 89, 31, 17, 127, 31, 3221, 179, 151, 127, 8191, 13, 262657, 11939, 199, 257, 127, 127, 337, 257, 524287, 73, 1093, 547, 137, 241, 1093, 547, 331, 131071, 1093, 599479, 8191, 137, 127, 8191
Offset: 1
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f[1] = 2; f[n_] := f[n] = DivisorSigma[1, f[n - 1]]; Table[FactorInteger[f[n]][[-1, 1]], {n, 50}] (* Amiram Eldar, Sep 23 2020 *)
A335534
a(n) = tribonacci(n) modulo Fibonacci(n).
Original entry on oeis.org
0, 0, 1, 2, 4, 7, 0, 3, 10, 26, 60, 130, 38, 173, 485, 175, 977, 273, 2789, 2065, 336, 15149, 22718, 39800, 5226, 54214, 2323, 251416, 418400, 93831, 977776, 1518664, 261912, 5208104, 2557037, 3549042, 21177270, 11203146, 36247269, 87596844, 44950918, 261069681
Offset: 1
For n=10, since tribonacci(10)=81 and Fibonacci(10)=55, a(10)=81 modulo 55 = 26.
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a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1, 3] mod (<<0|1>, <1|1>>^n)[1, 2]:
seq(a(n), n=1..45); # Alois P. Heinz, Aug 19 2020
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m = 42; Mod[LinearRecurrence[{1, 1, 1}, {0, 1, 1}, m], Array[Fibonacci, m]] (* Amiram Eldar, Aug 19 2020 *)
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t(n) = ([0, 1, 0; 0, 0, 1; 1, 1, 1]^n)[1, 3]; \\ A000073
a(n) = t(n) % fibonacci(n); \\ Michel Marcus, Aug 19 2020
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