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An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. There are two possibilities. For the first kind, the main diagonal is 0's=A000004, the first two following diagonals being the same (generally not A000004). Integers example: A000045(n).

For the second kind, the main diagonal is the double of the following diagonal. Example: the companion to A000045(n) is A000032(n)=2, 1, 3, ... .

A000032(n)/2 is also a possibility. Here a(n) is the denominator of the sum of two autosequences of second kind involving (fractional) Euler and Bernoulli numbers. The corresponding fractional sequence is also an autosequence of the second kind: 2, 1, 1/6, -1/4, -1/30, 1/2, 1/42, -17/8, -1/30, 31/2, 5/66, -691/4, -691/2730,... . It could be divided by 2.

There exist an infinity of 1's, 2's, 6's, 30's, 42's, 66's, ... .

Respective ranks:

0, 3, 7, 11, 15, 19, ...

1, 5, 9, 13, 17, 21, ... (= A016813)

2, 14, 26, 34, 38, 62, ... (= A051222)

4, 8, 68, 76, 124, 152, ... (= A051226)

6, 114, 186, 258, 354, 402, ... (= A051228)

10, 50, 170, 370, 470, 590, ... (= A051230)

12, 24, 1308, 1884, 2004, 2364, ... (= A249134)

etc.

Hence by antidiagonals a permutation of A001477(n).

First column: A248614(n).

a(n) is an alternative sequence for the denominators of the Bernoulli numbers.

First 36 terms of the corresponding clockwise spiral:

.

330------2----138------1---2730------2

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1 42------1-----30------2 6

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798 2 1------2 66 1

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2 30------1------6 1 870

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510------1------6------2---2730 2

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1------6------2----510------1--14322

Fractions: 1/2, 1/2, -4/3, 2, -38/15, 3, -73/21, 4, -68/15, 5, -179/33, 6, -9218/1365, 7, ... .

Numerators are A307974(n).

a(2n) same as denominators of cosecant numbers A001897 for n>0 (conjectured).

Denominators: 2, 2, 3, 1, 15, 1, 21, 1, 15, 1, 33, 1, ... .

Denominators 3, 15, 21, 15, 33, 1365, 3, 255, ... coincide with cosecant numbers A001897, except 1 (conjectured).

Squarefree composites m such that LCM_{prime p|m} (p-1) = LCM_{prime p, p-1|m-1} (p-1).

Carmichael numbers m such that LCM_{prime p|m} (p-1) = LCM_{prime p, p-1|m-1} (p-1), i.e., with A173614(m) = A346467(m).

Carmichael numbers m such that their index (m-1)/lambda(m) = A346468(m), cf. A174590.

Carl Pomerance noted that, for k = 40826, Chernick's Carmichael number (6k+1)*(12k+1)*(18k+1) = 88189878776579929 satisfies this condition.

Theorem: lambda(m) | lambda(D_{m-1}) if and only if m | D_{m-1}.

Composites m such that lambda(m) | lambda(D_{m-1}) are all Carmichael numbers, defined as composites m such that lambda(m) | m-1, while lambda(D_{m-1}) | m-1 for every m.

Note that if p is prime, then lambda(p) = lambda(D_{p-1}) = p-1.

In A190339 we saw that (the second Bernoulli numbers) A164555/A027642 is an eigensequence (its inverse binomial transform is the sequence signed) of the second kind, see A192456/A191302. We consider this array preceded by 1 for the second row, by 1, -3/2, for the third one; 1 is chosen and is followed by the differences of successive rows.

Hence

1 1/2 1/6 0

1 -1/2 -1/3 -1/6 -1/30

1 -3/2 1/6 1/6 2/15 1/15

1 -5/2 5/3 0 -1/30 -1/15 -8/105.

The second row is A051716/A051717.

The (reduced) triangle before the square array (T(n,m) in A190339) is a(n)/b(n)=

B(0)= 1 = 1 Redbernou1li

B(1)= -1/2 = 1 -3/2

B(2)= 1/6 = 1 -5/2 5/3

B(3)= 0 = 1 -7/2 25/6 -5/3

B(4)=-1/30 = 1 -9/2 23/3 -35/6 49/30

B(5)= 0 = 1 -11/2 73/6 -27/2 112/15 -49/30.

For the main diagonal, see A165142.

Denominator b(n) will be submitted.

This transform is valuable for every eigensequence of the second kind. For instance Leibniz's 1/n (A003506).

With increasing exponents for coefficients, polynomials CB(n,x) create Redbernou1li. See the formula.

Triangle Bernou1li for A027641/A027642 with the same denominator A080326 for every column is

1

1 -3/2

1 -5/2 10/6

1 -7/2 25/6 -10/6

1 -9/2 46/6 -35/6 49/30

1 -11/2 73/6 -81/6 224/30 -49/30.

For numerator by columns,see A000012, -A144396, A100536, Q(n)=n*(2*n^2+9*n+9)/2 , new.

Triangle Checkbernou1 with the same denominator A080326 for every row is

1/1

(2 -3)/2

(6 -15 +10)/6

(6 -21 +25 -10)/6

(30 -135 +230 -175 +49)/30

(30 -165 +365 -405 +224 -49)/30;

Hence for numerator: 1, 2-3, 16-15, 31-31, 309-310, 619-619, 8171-8166.

Absolute sum: 1, 5, 31, 62, 619, 1238, 17337. Reduced division by A080326:

1, 5/2, 31/6, 31/3, 619/30, 619/15, 5779/70, = A172030(n+1)/A172031(n+1).

Conjecture: a(15) = a(16) = 131070, a(17) through a(32) = 8589934590.

Number of different terms: 1, 1, 1, 2, 4, ... = abs(A141531)?

Factorization of terms from 2:

2 = 2

6 = 2*3

30 = 2*3*5

510 = 2*3*5*17

131070 = 2*3*5*17*257

8589934590 = 2*3*5*17*257*65537.

Note that all factors shown are 2 or Fermat numbers (see A092506, A019434, A000215).

Empirical: using the von Staudt-Clausen theorem, terms a(17) through a(4215) are all 8589934590. - Simon Plouffe, Sep 20 2015

Using the von Staudt-Clausen theorem, a(n) is the product of 2 and all Fermat primes <= 2^(n-1)+1: see A019434. The only known Fermat primes are 3,5,17,257,65537; it is known that there are no others < 2^(2^33)+1, so that a(n) = 8589934590 for n <= 2^33 = 8589934592. - Robert Israel, Sep 21 2015

Starting from any sequence a(k) in the first row, we define the array T(n,k) of the inverse bi-binomial transform by T(0,k) = a(k), T(n,k) = T(n-1,k+1) -2*T(n-1,k) n>0. Hence A164558(n)/A027642(n) and successive "bi-differences":

1, 3/2, 13/6, 3, 119/30, 5, 253/42, 7, 239/30, 9;

-1/2, -5/6, -4/3, -61/30, -44/15, -167/42, -106/21, -181/30, -104/15;

1/6, 1/3, 19/30, 17/15, 397/210, 61/21 , 853/210, 77/15;

0, -1/30, -2/15, -79/210, -92/105, -367/210, -314/105;

-1/30, -1/15, -23/210, -13/105, 1/210, 53/105;

0, 1/42, 2/21, 53/210, 52/105;

1/42, 1/21, 13/210, -1/105;

0, -1/30, -2/15;

-1/30, -1/15;

0.

The first column is A027641(n)/A027642(n).

r(n): 0, 0, 1, 1/2, 5/3, 13/6, 19/5, 179/30, 1028/105, 1103/70, 893/35,... = A227500(n)/a(n). a(0)=a(1)=1 is a choice.